IN MEMORIAM
FLOR1AN CAJORI
THE ONTARIO
HIGH SCHOOL GEOMETRY
THEORETICAL
BY
A. H. McDOUGALL, B.A.
PRINCIPAL OTTAWA COLLEGIATE INSTITUTE
Authorized by the Minister of Education for Ontario
TORONTO
THE COPP, CLARK COMPANY, LIMITED
Copyright, Canada, 1910, by THE COPP, CLARK COMPANY, LIMITED.
Toronto, Ontario.
FIRST EDITIOH, 1910.
RBPRINTED, 1911. 1912, 1913, 1914.
PREFACE
The Ontario High School Geometry is intended to cover
the course in Theoretical Geometry, begun in the Lower
School and completed in the Middle School, as defined in
the Programme of Studies for High Schools and Collegiate
Institutes of the Province of Ontario.
In deference to the wish of the teachers of mathematics
of the Province, this Geometry is divided into Books with
numbered propositions.
While the theoretical course is complete in itself, it is
assumed that its study has been preceded by the usual course
in drawing and measurement. A considerable number of
practical problems are given in the exercises. These should
be worked out carefully, and, in fact, all diagrams should be
accurately and neatly made.
The book contains an abundant supply of carefully selected
and graded exercises. Those given in sets throughout the
Books will be found suitable for the work of average classes,
and just about sufficient in number to fix the subject-matter
of the propositions in the minds of the pupils. All the
problems contained in the miscellaneous collections at the
ends of the Books could be worked through by a few of the
best pupils only, and should be used also by the teachers
as a store from which to draw suitable material for review
purposes from time to time.
While the requirements of class-work have been constantly
kept in mind in the choice of proofs, it should not be assumed
that other proofs, just as good, cannot in many cases be given.
iii
iv PREFACE
Students should be constantly encouraged to work out
methods of their own, and to keep records of the best in
their note books.
Symmetry has been used to an unusual extent in giving a
more concise form to the proofs of constructions.
The treatment of parallels, in accord with the method of
many of the best English text-books, is based on Playfair's
Axiom.
Tangents are treated both by the method of limits and as
lines which meet the circle in only one point.
Areas of triangles and parallelograms are compared with
rectangles, thereby not only giving a simple method of treat-
ment, but also promoting facility in numerical computations.
Similarly, the treatment of proportion is correlated with the
algebraic knowledge of the pupil.
OTTAWA, June, 1910.
SYMBOLS AND ABBREVIATIONS
The following symbols and abbreviations are used : —
Fig. Figure.
Const. Construction.
Hyp. Hypothesis.
Cor. Corollary.
e.g. exempli gratia, for example.
i.e. id est, that is.
p. page.
V because, since.
.'. therefore,
rt. right,
st. straight.
z_, z_s, <Ld angle, angles, angled.
A, As triangle, triangles.
||, || s parallel, parallels.
|| gm, ||gms parallelogram, parallelograms,
sq., sqs. square, squares.
AB2 the square on AB.
rect. rectangle.
AB.CD the rectangle contained by AB and CD.
AB : CD, or — the ratio of AB to CD.
+ plus, together with.
- minus, diminished by.
_L is perpendicular to, a perpendicular.
= is equal to, equals.
^> is greater than.
<[ is less than.
EE is congruent to, congruent.
Ill is similar to, similar,
v
CONTENTS
BOOK I PAQK
PRELIMINARY DEFINITIONS AND EXPLANATIONS . . 1
GEOMETRICAL REASONING 3
ANGLES AND TRIANGLES 8
FIRST CASE OP THE CONGRUENCE OP TRIANGLES . . 16
SECOND CASE OF THE CONGRUENCE OF TRIANGLES . 22
CONSTRUCTIONS 25
PARALLEL STRAIGHT LINES 35
TRIANGLES 45
THIRD CASE OP THE CONGRUENCE OF TRIANGLES. . 54
THE AMBIGUOUS CASE IN THE COMPARISON OF
TRIANGLES 56
INEQUALITIES 59
PARALLELOGRAMS 66
CONSTRUCTION 74
Loci 75
MISCELLANEOUS EXERCISES 83
BOOK II
AREAS OF PARALLELOGRAMS AND TRIANGLES . . , 93
CONSTRUCTIONS 110
AREAS OF SQUARES 116
MISCELLANEOUS EXERCISES 134
vii
Vlll CONTENTS
BOOK in PAOB
THE CIRCLE 141
CONSTRUCTIONS 143
ANGLES IN A CIRCLE 152
TANGENTS AND CHORDS . . . • 169
CONSTRUCTION 173
. ANGLE BETWEEN CHORD AND TANGENT ..... 177
CONSTRUCTIONS 182
CONTACT OF CIRCLES 196
MISCELLANEOUS EXERCISES 200
BOOK IV
RATIO AND PROPORTION . 213
CONSTRUCTIONS . . 226
BISECTOR THEOREMS 230
SIMILAR TRIANGLES 236
GEOMETRIC MEANS 244
RECTANGLES 247
PROOF OF PYTHAGOREAN THEOREM BY PROPORTION . 249
CHORDS AND TANGENTS 252
MISCELLANEOUS EXERCISES . 258
BOOK V
AREAS OF SIMILAR FIGURES . ....'. • • ... • 271
CONSTRUCTIONS 274
AREAS OF SIMILAR POLYGONS 278
CONSTRUCTIONS 280
ARCS AND ANGLES 284
ANALYSIS OF A PROBLEM — COMMON TANGENTS . . 285
MISCELLANEOUS EXERCISES * 289
INDEX 299
THEORETICAL GEOMETRY
BOOK I
PRELIMINARY DEFINITIONS AND EXPLANATIONS
1. A point is that which has position but no size.
The position of a point on the blackboard, or on
paper, is represented by a mark. This mark has
some small size and therefore only roughly represents
the idea of a point.
2. A line is that which has length but neither
breadth nor thickness.
Again, the mark that we use to represent a line
has breadth and some small thickness, and conse-
quently, only roughly represents the idea.
The intersection of two lines is a point.
3. Lines may be either straight or curved.
The following property distinguishes straight lines
from curved lines and may be used as the definition
of a straight line : —
Two straight lines cannot have any two points of
one coincide with two points of the other without
the lines coinciding altogether.
This is sometimes stated as follows: — Joining two
points there is always one and only one straight lina
1
2 THEORETICAL GEOMETRY ^ BOOK I
It follows from this definition that two straight
lines cannot enclose a space.
Can the circumferences of two equal circles coincide
in two points without coinciding altogether ?
4. A surface is that which has length and breadth
but no thickness.
A sheet of tissue paper has length and breadth
and very little thickness. It thus roughly represents
the idea of a surface. In fact the sheet of paper has
two well-defined surfaces separated by the substance
of the paper.
The boundary between two parts of space is a
surface.
5. Surfaces may be either plane or curved.
The following property distinguishes plane surfaces
from curved surfaces and may be used as the defini-
tion of a plane surface : —
The straight line joining any two points on a
plane surface lies wholly on that surface.
Give examples of curved surfaces on which straight
lines may be drawn in certain directions. Notice the
force of the word " any " in the definition above.
6. A solid is that which has length, breadth and
thickness.
7. Any combination of points, lines, surfaces and
solids is called a figure.
8. Geometry is the science which investigates the
properties of figures and the relations of figures to
one another.
GEOMETRICAL REASONING 3
9. In Plane Geometry the figure, or figures, con-
sidered in each proposition are confined to one plane,
while Solid Geometry treats of figures the parts of
which are not all in the same plane.
Plane Geometry is also called Geometry of Two
Dimensions (length and breadth), and Solid Geometry
is called Geometry of Three Dimensions (length,
breadth and thickness).
GEOMETRICAL REASONING
10. Two general methods of investigating the pro-
perties or relations of figures may be distinguished
as the Practical Method and the Theoretical Method.
Some properties may be tested by measurement,
paper-folding, etc., while in the same or other cases it
may be shown that the property follows as a neces-
sary result from others that are already known to be
true.
The Theoretical Method, has certain advantages over
the Practical method. Measurements, etc., are never
exact, and in many cases cannot be made directly; but
in the Theoretical Method, starting from certain simple
statements, called axioms, the truth of which is self-
evident, or, it may be in some cases, assumed, the
consequent statements follow with absolute certainty.
The Practical Method is also known as the Induc-
tive Method of Reasoning, and the Theoretical Method
as the Deductive Method.
4 THEORETICAL GEOMETRY BOOK I
11. Figures may be compared by making a tracing
of one of them and fitting the tracing on the other.
In many cases the process may be made a mental
operation and the comparison made with absolute
certainty by means of the following axiom : —
A figure may be, actually or mentally, transferred
from one position to another without change of form
or size.
When two figures are shown to be exactly equal in
all respects by supposing one to be made to fit exactly
on the other, the proof is said to be by the method
of superposition.
Figures which exactly fill the same space are said
to coincide with each other.
12. In general a proposition is that which is stated
or affirmed for discussion.
In mathematics a proposition is a statement of
either a truth to be demonstrated or of an operation
to be performed. It is called a theorem when it is
something to be proved, and a problem when it is
a construction to be made.
Example of Theorem: — If two straight lines cut
each other, the vertically opposite angles are equal.
Example of Problem : — It is required to bisect a
given straight line.
13. Theorems are commonly stated in two ways: —
First, the General Enunciation, in which the pro-
perty is stated as true for all figures of a class, but
without naming any particular figure, as in the first
example given in § 12; second, the Particular
Enunciation, in which the theorem is stated to be
true of the particular figure in a certain diagram.
GEOMETRICAL REASONING
Similarly general and particular enunciations are
commonly given for problems.
Examples of Particular Enunciation: —
1. Let AB and CD be two st. lines cutting at E.
It is required to show that Z AEG = Z BED, and
that L AED - L BEC.
2. Let AB be a given st. line.
It is required to bisect AB.
14. In general, the enunciation of a theorem con-
sists of two parts: the hypothesis and the conclusion.
The hypothesis is the formal statement of the
conditions that are supposed to exist, e.g., in the first
example of § 12, "If two straight lines cut each
other."
The conclusion is that which is asserted to follow
necessarily from the hypothesis, e.g., "the vertically
opposite angles are equal to each other."
Commonly, the hypothesis of a theorem is stated
first, introduced by the word " if," and the two parts
hypothesis and conclusion are separated by a comma,
Sometimes, however, the two parts are not so formally
6 THEORETICAL GEOMETRY BOOK I
distinguished, e.g., in the proposition : — The angles at
the base of an isosceles triangle are equal to each
other. In order to show the two parts, this statement
may be changed as follows: — If a triangle has two
sides equal to each other, the angles opposite these
equal sides (or angles at the base) are equal to each
other.
15. The demonstration of a theorem depends either
on definitions and axioms, or on other theorems that
have been previously shown to be true.
The following are some of the axioms commonly
used in geometrical reasoning: —
1. Things that are equal to the same thing are
equal to each other.
If A = B, B = C, C = D, D = E and E = F, what
about A and F?
2. If equals be added to equals the sums are
equal.
Thus if A, B, C, D be four at. lines such that
A = B and C = D, then the sum of A and C = the
sum of B and D.
Exercise : — Mark four successive points A, B, C, D
on a st. line such that AB = CD. Show that AC = BD.
3. If equals be taken from equals the re-
mainders are equal.
Give example.
GEOMETRICAL REASONING 7
Exercise : — Mark four successive points A, B, C, D
on a st. line such that AC = BD. Show that AB = CD.
4. If equals be added to unequals the sums are
unequal, the greater sum being obtained from the
greater unequal.
Give example. Show also, by example, that if
unequals be added to unequals the sums may be
either equal or unequal.
5. If equals be taken from unequals the re-
mainders are unequal, the greater remainder being
obtained from the greater unequal.
6. Doubles of the same thing, or of equal
things, are equal to each other.
7. Halves of the same thing, or of equal things,
are equal to each other.
8. The whole is greater than its part, and
equal to the sum of all its parts.
Give examples.
9. Magnitudes that coincide with each other, are
equal to each other.
These simple propositions, and others that are also
plainly true, may be freely used in proving theorems.
THEORETICAL GEOMETRY
ANGLES AND TRIANGLES
BOOK 1
16. Definitions. — When two straight lines are
drawn from a point they are said to form an angle.
The point from which the two lines are drawn is
called the vertex of the angle.
The two lines are called the arms of the angle.
The angle in the figure may be called the angle
BAG, or the angle CAB. The letter at the vertex
must be the middle one in reading the angle.
The single letter at the vertex is sometimes used to
denote the angle when there can be no doubt as to
which angle is meant.
17. Suppose a straight line OB to be fixed, like a
rigid rod on a pivot at the point O, and be free to
rotate in the plane of the paper.
If the line OB start from any position OA, it may
rotate in either of two directions — that in which the
hands of a clock rotate, or in the opposite.
ANGLES AND TRIANGLES 9
When OB starts from OA and stops at any position
an angle is formed with O for its vertex and OA and
OB for its arms.
18. An angle is said to be positive or negative
according to the direction in which the line that
traces out the angle is supposed to have rotated.
The direction contrary to that in which the hands of
a clock rotate is commonly taken as positive.
19. The magnitude of an angle depends altogether
on the amount of rotation, and is quite independent
of the lengths of its arms.
20. If we wish to compare two angles ABC and
DEF we may suppose the angle ABC to be placed on
the angle DEF so that B falls on E and BA along
ED. The position of BC with respect to EF will then
show which of the angles is the greater and by how
much it is greater than the other.
21. Definition. — When a revolving line OB has made
half of a complete revolution from the initial position
OA the angle formed is a straight angle.
The arms of a straight angle are thus in the same
straight line and extend in opposite directions from
10 THEORETICAL GEOMETRY BOOK I
the vertex. At the point O, in the diagram, there are
two straight angles on opposite sides of the straight
line AOB, the two straight angles making up the com-
plete revolution.
22. Definition. — If a straight line, starting from OA,
rotates in succession through two equal angles AOB,
BOG, the sum of which is a straight angle, each of
these angles is called a right angle.
A right angle is thus one-half of a straight angle,
or one-quarter of a complete revolution.
Each arm of a right angle is said to be perpen-
dicular to the other arm.
What is a vertical line ? a horizontal line ?
An angle which is less than a right angle is called
an acute angle.
An angle which is greater than a right angle is
called an obtuse angle.
23. If a right L be divided into ninety equal
parts, each of these parts is called a degree.
Thus 1 rt. L m 90°,
1 st. L = 180°
1 revolution = 360°.
ANGLES AND TRIANGLES 11
24. Let a st. line starting from OA revolve through
two successive z_s
AOB, BOG such that OC is in the same st. line with
OA, but in the opposite direction from the point O,
and consequently AOC is a st. L.
•: L AOB + Z BOC = the st. Z AOC,
;. Z AOB 4- Z BOC = 2 rt. Zs.
Thus the angles which one straight line makes
with another on the same side of that other are
together equal to two right angles.
25. Definition. — When two angles have the same
vertex and a common arm, and the remaining arms
on opposite sides of the common arm, they are said
to be adjacent angles.
A!
Thus BAG and CAD are adjacent angles having the
same vertex A and the common arm AC.
But angles BAD and CAD, with the same vertex
and the common arm AD are not adjacent angles.
12
THEORETICAL GEOMETRY
BOOK I
26. Let the adjacent Ls ABC, ABD be together
equal to two rt. z_s.
/A
L ABD + L ABC = two rt. Z_s = a st. L.
That is, L DBC is a st. Z_,
and .'. line DBC is a st. line.
Thus, if two adjacent angles are together equal
to two right angles, the exterior arms of the
angles are in the same straight line.
27. Let a st. line OB, starting from the position
OA, and rotating in the positive direction, trace out
the successive Ls: AOC, COD, DOE, EOF, FOA.
The sum of the successive Ls is a complete revolu-
tion, and therefore equal to four rt. La.
Thus, if any number of straight lines meet at
a point, the sum of the successive angles is four
right angles.
ANGLES AND TRIANGLES 13
THEOREM 1
Each of the angles formed by two intersecting
straight lines is equal to the vertically opposite
angle.
B
Hypothesis. — The two st. lines AB, CD cut each
other at E.
To prove that (1) L AEC = L BED,
(2) L AED = L BEC.
Proof. — V CED is a st. line,
L AEC + L AED = two rt. Zs.
AEB is a st. line,
L AED -f L DEB = two rt. Z_s.
/. Z AEC -f Z AED = Z AED 4- L DEB.
From each of these equals take away the common
Z. AED and the remainders must be equal to each
other.
/. L AEC = Z DEB.
In the same manner it may be shown that Z. AED
= L CEB.
28. Definitions. — When two angles are such that
their sum is two right angles, they are said to be
supplementary angles, or each angle is said to be the
supplement of the other.
If two L s are equal, what about their supple-
mentary L S ?
When two angles are such that their sum is one right
angle, they are said to be complementary angles, or each
angle is said to be the complement of the other.
14
THEORETICAL GEOMETRY
BOOK I
. 29. — Exercises
1. If one of the four L s made by two intersecting st.
lines be 17°, find the number of degrees in each of the
other three.
2. Two st. lines ABD, CBE cut at B, and Z ABC is a
rt. L . Prove that the other L. s at B are also rt. L. s.
3. If in the figure of Theorem 1 the
find the number of degrees in each L. of the figure.
4.
DOC is a rt. L , and through the
vertex O a st. line AOB is drawn.
Prove that : —
In Fig. 1, Z BOG -f Z AOD - a rt. L.
In Fig. 2, Z BOG - Z AOD = a rt. L. .
5. In the diagram,
Z ABC = Z ACB.
Prove that
(1) Z ABD = Z ACE,
(2) Z FBC = Z GCB,
(3) Z DBF = Z EGG.
6. In the diagram,
AOB is a st. line,
Z COD = Z DOB and
Z AOE = Z EOC.
Prove that EOD is a rt. L , and that Z AOE is the com-
plement of Z BOD.
7. E is a point between A and B in the st. line AB; DE, FE
are drawn on opposite sides of AB and such that Z DEA =
Z FEB. Show that DEF is a st. line.
ANGLES AND TRIANGLES 15
8. Four st. lines, OA, OB, OC, OD, are drawn in succes-
sion from the point O, and are such that Z AOB = Z COD
and Z BOC = Z DOA. Show that AOC is a st. line, and
also that BOD is a st. line. ^
A__ B/ C
9. In the diagram, ABC, DEF,
GBEH are st. lines and Z ABE =
Z BEF. V—
Prove that H
(1) Z CBE = Z BED,
(2) Z GBC = Z DEH,
(3) Z ABG = Z BED,
(4) Zs CBE, BEF are supplementary,
(5) Zs ABE, BED are supplementary.
30. Definitions. — A figure formed by straight lines
is called a rectilineal figure.
The figure formed by three straight lines which
intersect one«another is called a triangle.
The three points of intersection are called the
vertices of the triangle.
The lines between the vertices of the triangle are
called the sides of the triangle.
31. Figures that are equal in all respects, so that
one may be made to fit the other exactly, are said to
be congruent
The sign = is used to denote the congruence of
figures.
16 THEORETICAL GEOMETRY BOOK I
FIRST CASE OF THE CONGRUENCE OF TRIANGLES
THEOREM 2
If two triangles have two sides and the contained
angle of one respectively equal to two sides and the
contained angle of the other, the two triangles are
congruent.
Hypothesis. — ABC and DBF are two As having
AB = DE, AC = DF and L A = L D.
To prove that (1) BC = EF,
(2) L B = L E,
(3) L C = L F,
(4) area of A ABC == area of A DEF;
and, hence, A ABC = A DEF.
Proof. — Let A ABC be applied to A DEF so that
vertex A falls on vertex D and AB falls along DE.
V AB = DE.
.-. vertex B must fall on vertex E. .
V L A = L D,
/. AC must fall along DF,
and .-. , as AC = DF,
the vertex C must fall on the vertex F.
/. A ABC coincides with A DEF.
and /. A ABC = A DEF.
FIRST CASE OF THE CONGRUENCE OF TRIANGLES 17
32. Definitions. — A closed figure formed by four
straight lines is called a quadrilateral.
In a quadrilateral a straight line joining two oppo-
site vertices is called a diagonal.
A quadrilateral having its four sides equal to each
other is called a rhombus.
A circle is a figure consisting of one closed curved
line, called the circumference, and is such that all
straight lines drawn from a certain point within the
figure, called the centre, to the circumference are
equal to each other.
In a circle a st. line drawn from the centre to the
circumference is called a radius. (Plural — radii.)
A st. line, as AB, joining two points in the circum-
ference is called a chord.
If a chord passes through the centre, as GD, it is
called a diameter.
A part of the circumference, as the curved line
FED, is called an arc.
A line drawn from a point in one arm of an angle
to a point in the other arm is said to subtend the
angle. In the diagram the arc FE subtends the L
FCE; or in any A each side subtends the opposite L.
18 THEORETICAL GEOMETRY BOOK I
33.— Exercises
1. Prove Theorem 2 when one A has
to be supposed to be turned over before
it can be made to coincide with the
other.
2. The L B of a A ABC is a rt. L, and
CB is produced to D making BD = BC. Prove AD = AC.
3. A, B, C are three points in a st. line such that
AB = BC. DB is J_ AC. Show that any point in DB,
produced in either direction, is equidistant from A and C.
4. Two st. lines AOB, COD cut one another at O, so that
OA = OB and OC = OD; join AD and BC, and prove As
AOD, BOG congruent.
5. Prove that all chords of a circle which subtend equal
angles at the centre are equal to each other.
6. If with the same centre O, two circles be drawn, and
st. lines ODB, OEC be drawn to meet the circumferences in
D, E, B, C; prove that BE = DC.
7. ABCD is a quadrilateral having the opposite sides
AB, CD equal and L B = Z C. Show that AC = BD.
8. In the diagram, ABC and DEF are A B . C
both i. BE. Also AB = BC and DE = 71
EF. Prove that AD = CF. /
9. Two st. lines AOB, COD cut one **
another at rt. /_s at O. AO is cut off = OB, and CO =
OD. Prove that the quadrilateral ACBD is a rhombus.
10. Two quadrilaterals ABCD, EFGH have AB = EF,
BC = FG, CD = GH, Z B = Z F, Z C = Z G. Prove that
they are congruent.
FIRST CASE OF THE CONGRUENCE OF TRIANGLES 19
34. Definitions. — A triangle having its sides all equal
to each other is called an equilateral triangle.
A triangle having two sides equal to each other is
called an isosceles triangle.
A triangle having no two of its sides equal to each
other is called a scalene triangle.
35.
If a straight line revolve in the positive direction
about the point O from the position OA to the position
OB, it must pass through some position OC such that
L AOC = L COB.
A straight line which divides an angle into two
equal angles is called the bisector of the angle.
When a construction is represented in a diagram,
although it has not previously been proved that it
can be made, it is called a hypothetical construction.
Thus OC has been drawn to represent the bisector of
L AOB.
20 THEORETICAL GEOMETRY BOOK I
THEOREM 3
The angles at the base of an isosceles triangle
are equal to each other.
B D C
Hypothesis. — ABC is an isosceles A having AB = AC.
To prove that L B = L C.
Hypothetical Construction. — Draw the st. line AD to
represent the bisector of L BAG.
Proof. — In the two As ADB, ADC,
AB = AC, (Hyp.)
AD is common,
Z BAD = Z CAD, (Const.)
A ADB = A ADC, (1—2, page 16.)
Z B = Z C.
36. The two As ADB, ADC, in the diagram of Theo-
rem 3, are congruent, and if the isosceles A be folded
along the bisector of the vertical L as crease, the parts
on one side of the bisector will exactly fit the corre-
sponding parts on the other side.
Definition. — When a figure can be folded along a
line so that the part on one side exactly fits the part
on the other side, the figure is said to be symmetrical
with respect to that line.
EXERCISES 21
The line along which the figure is folded is called
an axis of symmetry of the figure.
Hence the bisector of the vertical L of an isos-
celes A is an axis of symmetry of the A.
It follows from the above definition of a symmetrical
figure that —
If a figure is symmetrical with respect to a st.
line, for every point on one side of this axis of
symmetry there is a corresponding point on the
other side.
Show by folding, in the diagram of Theorem 3,
that if Z B = L C, the side AB = the side AC.
37.— Exercises
1. An equilateral A is equiangular.
2. ABC is an equilateral A, and points D, E, F, are
taken in BC, CA, AB respectively, such that BD = CE = AF.
Show that DEF is an equilateral A.
3. Show that the exterior z_s at the base of an isosceles
A are equal to each other.
4. The opposite Z_s of a rhombus are equal to each other.
5. ABC is an isosceles A having
AB = AC, and the base BC pro-
duced to D and E such that BD
= CE. Prove that ADE is an
isosceles A. £) "6 c11 EL
6. AC, AD are two st. lines on opposite sides of AB.
Prove that if the bisectors of /_s BAG, BAD are at rt. Zs,
AC, AD must be in the same st. line.
7. If a figure be symmetrical with respect to a st. line,
the st. line joining any two corresponding points cuts the
axis at rt. /_s.
22 THEORETICAL GEOMETRY BOOK I
SECOND CASE OF THE CONGRUENCE OF TRIANGLES
THEOREM 4
If two triangles have the three sides of one
respectively equal to the three sides of the other,
the two triangles are congruent.
Hypothesis. — ABC, DEF are two As having AB = DE,
AC = DF and BC = EF.
To prove that A ABC = A DEF.
Proof. — Let A DEF be applied to A ABC so that the
vertex E falls on the vertex B and EF falls along BC.
Then v EF = BC, the vertex F falls on C. Let D take
the position D' on the side of BC remote from A.
Join AD'.
BA = BD',
/. L BAD'= Z BD'A. (1—3, p. 20.)
Similarly Z CAD'= Z CD'A.
.'. Z_BAD' + Z CAD'- Z BD'A -f Z CD'A,
i.e., Z BAC= Z BD'C.
f BA=BD',
Then in As BAG, BD'c] CA = CD',
U BAC= Z BD'C,
/. A ABC E A BD'C; (1—2, p,
i.e., A ABC = A DEF.
16.)
SECOND CASE OF THE CONGRUENCE OF TRIANGLES 23
Note. — In the proof of this theorem three cases may
occur : — AD' may cut BC as in Fig. 1, or not cut BC as in
Fig. 2, or pass through one end of BC as in Fig. 3.
FIG. 1
PICK 2
FIG. 3
The proof given above is that of the first case. The
pupil should work out the proofs of the other two cases.
38.— Exercises
1. If the opposite sides of a quadrilateral be equal, the
opposite zs are equal.
2. A diagonal of a rhombus bisects each of the zs
through which it passes, and consequently, the diagonal is
an axis of symmetry in the rhombus.
3. If in a quadrilateral ABCD the sides AB, CD be equal
and Z ABC = Z BCD, prove that Z CDA = Z DAB.
4. Show that equal chords in a circle subtend equal zs
at the centre.
5. Prove that the diagonals of a rhombus bisect each
other at rt. zs.
24
THEORETICAL GEOMETRY
BOOK I
THEOREM 5
If two isosceles triangles are on the same base,
the straight line joining their vertices is an axis
of symmetry of the figure; and the ends of the
base are corresponding points.
C B
Hypothesis. — ABC, DBC are two isosceles As on the
same base BC.
To prove that AD is an axis of symmetry of the
figure.
Proof. — AD, or AD produced, cuts BC at E.
fAB = AC
BD = CD,
AD is common,
.-. A BAD = A CAD. (1—4, p. 22.)
and . . Z BAD *= Z CAD.
!BA t= CA,
AE is common,
Z BAE = Z CAE,
/. A BAE = A CAE. (1—2, p. 16.)
Similarly,
EXERCISES — CONSTRUCTIONS 25
Hence, each part of the figure on one side of AD is
congruent to the corresponding part on the other side,
and if the figure be folded on AD, as crease, the corre-
sponding parts will coincide.
.*. AD is an axis of symmetry of the figure; and
B, C are corresponding points.
39.— Exercises
1. If two circles cut at two points, the st. line which
joins their centres bisects at rt. zs the st. line joining the
points of section.
2. A, B, C are three points each of which is equidistant
from two fixed points P, Q. Show that A, B, C are in a
st. line which bisects the st. line joining P, Q and cuts it
at rt. /s.
CONSTRUCTIONS
40. In Theoretical Geometry the use of instruments
in making constructions is generally restricted to an
ungraduated straight edge and a pair of compasses.
With these instruments we can: —
1. Draw a st. line from one point to another.
2. Produce a st. line.
3. Describe a circle with any point as its centre
and radius equal to any given st. line.
4. Cut off from one st. line- a part equal to another
st. line.
NOTE. — All constructions should be accurately and neatly
drawn by the pupil, and, by means of theorems already
proved, the correctness of the method of construction should,
be shown.
26 THEORETICAL GEOMETRY BOOK I
PROBLEM 1
To bisect a given angle.
A
Let BAG be the given /_•
Construction. — With the compasses cut off equal
distances AD and AE from the arms of the z_.
With centre D describe an arc.
With centre E and the same radius describe another
arc cutting the first at F.
Join AF.
Then AF is the bisector of L BAG.
Proof. — Join DF, EF, DE.
ADE, FDE are isosceles As on the same base DE,
/. AF is an axis of symmetry of the figure, (I — 5, p. 24.)
/. AF bisects L BAG.
NOTE. — The equal radii fur the arcs with centres D and E
must be taken long enough for the arcs to intersect.
41. — Exercises
1. Divide a given / into four equal parts.
2. Prove that the bisectors of a pair of vertically op-
posite zs are in the same st. line.
3. Bisect a st. Z-
CONSTRUCTIONS — EXERCISES 27
PROBLEM 2
To draw a perpendicular to a given straight
line from a given point in the line.
Let CD be the given st. line and B the given point.
Construction. — Bisect the st. L CBD by the st. line
BG.
Proof. — Then each of the z.s CBG, DBG is half of a
st. L and /. each is a rt. L.
;. BG is J_ CD.
42. — Exercises
Using ruler and compasses only, construct ^-s of (1), 45° ;
(2), 221°; (3), 135°; (4), 671°; (5), 225°.
43. Definitions.- — If one angle of a triangle be a
right angle, the triangle is called a right-angled
triangle.
In a right-angled triangle the side opposite the right
angle is called the hypotenuse.
If one angle of a triangle be an obtuse angle, the
triangle is called an obtuse-angled triangle.
If all three angles of a triangle be acute angles, the
triangle is called an acute-angled triangle.
The altitude of a triangle is the length of the
perpendicular from any vertex to the opposite side.
28 THEORETICAL GEOMETRY BOOK I
44. — Exercises
1. Construct a rt.-/d A having one of the arms of the rt.
Z three times the other.
2. Construct a rt.-/d A having the hypotenuse three times
one of the arms of the rt. Z.
3. Given the length of the hypotenuse and of one of the
sides of a rt.-/d A, construct the A.
4. Construct a rhombus having each of its diagonals
equal to twice a given st. line.
5. Construct a rhombus having one diagonal twice and
the other four times a given st. line.
6. Construct an isosceles A having given its altitude and
the length of one of the equal sides.
7. Construct an isosceles rt.-Zd A.
45. Definitions. — Sometimes when a proposition has
been proved the truth of another proposition follows
as an immediate consequence of the former; such a
proposition is called a corollary.
A straight line which bisects a line of given length
at right angles is called the right bisector of the line.
CONSTRUCTIONS 29
PROBLEM 3
To bisect a given straight line.
X ! \
\
V>B
/
\
E
\
\
f*
X
/<
\
\
/
Let AB be the given st. line.
Construction. — With centre A and any radius that is
plainly greater than half of AB, draw two arcs, one
on each side of AB.
With centre B and the same radius draw two arcs
cutting the first two at C and D.
Join CD, cutting AB at E.
E is the middle point of AB.
Proof.— Join CA, AD, DB, BC.
CAB, DAB are isosceles As on the same base AB,
.-. CD is an axis of symmetry of the figure ; and
A, B are corresponding points. (I — 5, p. 24.)
.-. AE = EB.
Corollary. — From the above proof it follows that
the LS at E are rt. A.S, and hence, CD is the right
bisector of AB.
30
THEORETICAL GEOMETRY
BOOK I
46. Definition. — The straight line drawn from a
vertex of a triangle to the middle point of the oppo-
site side is called a median of the triangle.
47.— Exercises
1. Divide a given st. line into four equal parts.
2. In an isosceles A prove that the bisector of the verti-
""^cal ^ is a median of the A.
3. In an equilateral A prove that the bisectors of the Z s
are medians of the A.
f/"" 4. Show that any point in the right bisector of a given
st. line is equidistant from the ends of the given line.
/. 5. In any A the point of intersection of the right
* bisectors of any two sides is equidistant from the three
A. vertices.
If 6. The right bisectors of the three sides
of a A pass through one point.
The right bisectors of AB, BC meet at O.
Bisect AC at E. Join EO. Prove OE _L AC.
7. Describe a circle through the three
vertices of a A.
8. Describe a circle to pass through
three given points that are not in the
same st. line.
9. Show how any number of circles may
be drawn through two given points.
What line contains the centres of all
these circles?
10. In a given st. line find a point that is equally
distant from two given points.
11. On a given base describe an isosceles A so that the
sum of the two equal sides may equal a given st. line.
In what case is this impossible ?
CONSTRUCTIONS 31
12. Construct a rhombus having its diagonals equal to
two given at. lines. ,
* 13. In A ABC find in CA, produced if necessary, a point
D so that DC = DB.
^•14. In As ABC, DEF, AB = DE, AC = DF arid the
medians drawn from B and E are equal to each other.
Prove that A ABC = A DEF.
PROBLEM 4
To draw a perpendicular to a given straight line
from a given point without the line.
£
/\
A CN, .XD B
Let P be the given point and AB the given st. line.
Construction. — Describe an arc with centre P to cut
AB at C and D.
With centres C and D, and equal radii, describe two
arcs cutting at E.
Join PE, cutting AB at F.
PF is the required perpendicular.
Proof.— Join PC, CE, ED, DP.
/. PCD, ECD are isosceles As on the same base CD,
r .'. PE is an axis of symmetry of the figure; and
C, D are corresponding points. (I — 5, p. 24)
•*. z.s at F are rt. z.s, and PF is J_ AB.
32 THEORETICAL GEOMETRY BOOK I
PROBLEM 5
To construct a triangle with sides of given
length.
Let AB, C and D be the given lengths.
Construction. — With centre A and radius C describe
an arc.
With centre B and radius D describe an arc cutting
the first arc at E.
Join EA, EB.
AEB is the required A.
QUESTION — In what case would the above construction fail ?
48.— Exercises
1. On a given st. line describe an equilateral A.
2. On a given base describe an isosceles A having each
of the equal sides double the base.
3. Construct a rhombus having given a diagonal and the
length of one of the equal sides.
CONSTRUCTIONS 33
PROBLEM 6
To construct an angle equal to a given angle.
A IE c P
Let BAG be the given L.
Construction. — From AC, AB cut off equal parts
AE, AD.
Draw a line and mark a point P in it.
Cut off PQ = AE.
With centre P and radius PQ describe an arc.
With centre Q and radius DE describe an arc cutting
the arc with centre P at R.
Join RP.
RPQ is the required z_.
Proof. — Join DE, RQ.
f PQ - AE,
In As PRQ, ADE. J PR = AD,
[ RQ = DE,
:. z RPQ - z BAG. (l_4, p. 22.)
34 THEORETICAL GEOMETRY BOOK I
49.— Exercises
1. Construct a rhombus having given one of its Zs and
the length of one of its equal sides.
2. Construct a quadrilateral equal in all respects to a
given quadrilateral.
3. On a given st. line BC construct a A having the ' z s
B, C equal to two given acute zs.
4. Construct an / equal to the complement of a given
acute /.
5. Construct an / equal to the supplement of a given /.
6. On a given base describe an isosceles A having its
altitude equal to a given st. line.
7. In the side BC of a A ABC find a point E, such that
AE is half the sum of AB and AC.
8. The A formed by joining the middle points of the
three sides of an isosceles A is isosceles.
9. AB is a given st. line and C is a given point without
the line. Find the point D so that C and D may be
symmetrical with respect to AB.
10. C, D are given points, (1) on opposite sides, (2) on
the same side of a given st. line AB. Find a point P in
AB so that CP, DP make equal /s with AB.
11. The right bisectors of the two sides AB, AC of
A ABC meet at D, and E is the middle point of BC.
Show that DE _L BC.
PARALLEL STRAIGHT LINES 35
PARALLEL STRAIGHT LINES
50. Definitions. — Two straight lines in the same
plane which do not meet when produced for any
finite distance in either direction are said to be
parallel to each other.
A straight line which cuts two, or more, other
straight lines is called a transversal.
A quadrilateral that has both pairs of opposite
sides parallel to each other is called a parallelogram.
Draw a st. line EF cutting two other st. lines AB
and CD at G and H.
Eight z.s are thus formed, four of which, AGH, BGH,
CHG, DHG, being between AB and CD, are called
interior z_s. The other four are called exterior z_s.
The interior z_s AGH and GHD, on opposite sides of
the transversal, are called alternate z_s. Thus also,
BGH and GHC are alternate /_s.
Name four pairs of equal angles in the diagram.
36 THEORETICAL GEOMETRY BOOK I
THEOREM 6
If a transversal meeting two straight lines makes
the alternate angles equal to each other, the two
straight lines are parallel.
d;
c
_: W
D
r
HA
E
i
F
Hypothesis. — The transversal AB meeting CD and
EF makes L CGH = the alternate L GHF.
To prove that CD || EF.
Proof. — Detach the part DGHF from the figure and
mark it d g h f.
Slide d g h f, from its original position, along the
transversal until h comes to the point G.
Then, rotate d g A /, in either direction, through a
st. L about the point G.
When the rotation is complete h g coincides with
G H.
And, V L fhg = L CGH,
.*. h f coincides with GC.
Also, v L d g h = L GHE,
.*. g d coincides with HE.
PARALLEL STRAIGHT LINES 37
If it be possible let CD and EF when produced meet
towards D and F.
Then h f and g d must meet towards / and d,
:. GC and HE must meet towards C and E.
Hence, CD and EF when produced must meet in two
points.
This is impossible by the definition of a st. line.
/. CD and EF do not meet towards D and F, and
hence cannot meet towards C and E.
.'. CDIIEF.
NOTE. — If this proof is not at once clear to the
pupil he should make a drawing of the diagram, cut
out the part d g h f, and turning it about, Jit it to
E H G C.
51.— Exercises
1. Lines which are _L to the same ,st. line are || to each
other.
2. If both pairs of opposite sides of a quadrilateral are
equal to each other, the quadrilateral is a ||gm.
3. A rhombus is a ||gm.
^4. If the diagonals of a quadrilateral bisect each other,
ths quadrilateral is a ||gm.
_5. No two st. lines drawn from two vertices of a A, and
terminated in the opposite sides, can bisect each other.
38 THEORETICAL GEOMETRY BOOK I
THEOREM 7
If a transversal meeting- two straight lines makes
(i) an exterior angle equal to the interior and
opposite angle on the same side of the transversal,
or, (2) the two interior angles on the same side of
the transversal supplementary, in either case the
two straight lines are parallel.
-A
/
B'
(1) Hypothesis. — AB meeting CD, EF makes /_ AGD =
L GHF.
To prove. — CD || EF.
Proof.— Z CGH = Z AGO, (I— 1, p. 13.)
but Z AGO = Z GHF, (Hyp.)
:. Z CGH = Z GHF.
/. CD II EF. (1—6, p. 36.)
(2) Hypothesis. — AB meeting CD, EF makes Z DGH
+ Z GHF = two rt. L&.
To prove.— CD \\ EF.
Proof. — L CGH -f Z DGH = two rt. z_s,
but L DGH + Z GHF two rt. z_s, (Hyp.)
:. L CGH + Z DGH = Z DGH + Z GHF.
From each take the common L DGH, and L CGH =
L GHF,
:. CD || EF (1—6, p. 36.)
PARALLEL STRAIGHT LINES 39
52. The following statement of a fundamental pro-
perty of parallel straight lines is called Playfair's
axiom : —
Through any point one, and only one, straight
line can be drawn parallel to a given straight line.
From this axiom it follows that: —
No two intersecting straight lines can be parallel
to the same straight line.
/. straight lines which are parallel to the same
straight line are not intersecting lines, i.e. : —
Straight lines which are parallel to the same
straight line are parallel to each other.
40 THEORETICAL GEOMETRY BOOK I
THEOREM 8
If a transversal cuts two parallel straight lines,
the alternate angles are equal to each other.
c
K
C /
/B
Hypothesis. — The transversal AB cuts the || st. lines
CD, EF at G, H.
To prove that Z CGH = Z GHF.
Proof.— If Z CGH be not equal to Z GHF, make the
Z KGH = Z GHF, and produce KG to L.
Then Y AB cuts KL and EF, making Z KGH = the
alternate Z GHF.
/. KL is || to EF. (1—6, p. 36.)
But CD is, by hypothesis, || to EF.
That is, two intersecting st. lines, KL and CD, are
both [ EF, which is impossible.
.'. Z CGH * Z GHF.
53. Consider the method of proof used in Theorem 8.
To prove that Z CGH = Z GHF we began by assuming
that these Zs are not equal, and then showed that
something absurd or contrary to the hypothesis must
follow, and concluded that Z CGH = z GHF.
PARALLEL STRAIGHT LINES 41
This method of proof, in which we begin by assuming
that the conclusion is not true, is called the indirect
method of demonstration.
54. Compare Theorems 6 and 8.
In both cases a transversal cuts two straight lines.
In Theorem 6 the hypothesis is that the alternate
angles are equal, and the conclusion is that the lines
are parallel.
In Theorem 8 the hypothesis is that the lines are
parallel, and the conclusion is that the alternate angles
are equal.
Thus in these propositions the hypothesis of each is
the conclusion of the other.
When two propositions are such that the hypothesis
of each is the conclusion of the other, they are said to
be converse propositions ; or each is said to be the
converse of the other.
The converse of a true proposition may, or may not,
be true. The converse propositions in Theorems 6 and
8 are both true; but consider the true proposition:—
All rt. z_s are equal to each other; and its converse: —
All equal z_s are rt. Z_s. The last is easily seen to be
untrue. Consequently proof must in general be given
for each of a pair of converse propositions.
When a proposition is known to be true and we
wish to prove the converse we commonly use the
indirect method
42 THEORETICAL GEOMETRY BOOK I
THEOREM 9
If a transversal cuts two parallel straight lines, it
makes (i) an exterior angle equal to the interior
and opposite angle on the same side of the trans-
versal, and (2) the interior angles on the same side
of the transversal supplementary.
Hypothesis. — AB cuts the || st. lines CD, EF.
To prove that (1) Z AGO = Z AHF.
(2) Z DGH + Z GHF - two rt. Ls.
Proof.— (1) v CD || EF,
/. Z GHF = Z CGH. (1—8, p. 40.)
but Z CGH = Z AGO, (I— 1, p. 13.)
.'. Z AGO = Z GHF.
(2) v Z GHF = Z CGH,
.. Z GHF+ Z DGH = Z CGH -f Z DGH;
but Z CGH 4- Z DGH = a st. L
.*. Z_s GHF, DGH are supplementary.
PARALLEL STRAIGHT LINES 43
PROBLEM 7
Through a given point to draw a straight line
parallel to a given straight line.
AC D B
Let P be the given point and AB the given st. line.
Construction. — Take two points C, D, in AB.
With centre P and radius CD describe an arc.
With centre D and radius CP describe an arc cutting
the first at Q.
Join PQ.
Then PQ || AB.
Proof. — Join PC, DQ, PD.
f PC = DQ,
In As PCD, DQP, -j CD = Qp'
I PD is common,
/. Z CDP = Z DPQ. (1—4, p. 22.)
PQ || AB. (1—6, p. 36.)
55. —Exercises
I 1. If a st. line be JL to one of two || st. lines, it is also ±
to the other.
. I 2. Prove, by using a transversal, that st. lines which are ||
to the same st. line are II to each other.
44 THEORETICAL GEOMETRY BOOK I
/ 3. Any st. line || to the base of an isosceles A makes equal
/s with the sides, or the sides produced.
4. Construct a A having two of its /s respectively equal
to two given / s, and the length of the _L from the vertex
of the third Z to the opposite side equal to a given st. line.
5. Construct a rt.-/d A having given one side and the
opposite /.
6. If one Z of a ||gm be a rt. /, the other three /s are
also rt. /s.
7. Give a proof for the following method of drawing
a line through P || AB :—
Place the set-square with the hypotenuse along the st.
line AB.
Place a ruler against another side of the set-square as in
the diagram.
Hold the ruler firmly in position and slide the set-square
along it until the hypotenuse comes to the point P.
A line drawn through P along the set-square is |j AB.
ANGLES OF A TRIANGLE 45
TRIANGLES
THEOREM 10
The exterior angle, made by producing one side
of a triangle, equals the sum of the two interior
and opposite angles; and the three interior angles
are together equal to two right angles.
Hypothesis. — ABC is a A having BC produced to D.
To prove that (1) L ACD = L A+ z B.
(2) L A 4- Z B+ L ACB = two rt. £s.
Construction. — Through C draw CE || AB.
Proof. — •. CE il AB,
and AC is a transversal,
/. L ACE = L A. (1—8, p. 40.)
v BD is a transversal,
/. L ECD = L B. (1—9, p. 42.)
/. L ACE + L ECD = Z A+ Z B.
i.e., Z ACD = Z A-f Z B.
Hence, L A+ L B-f Z ACB = Z ACD+ Z ACB.
But Z ACD + Z ACB = two rt. Zs,
/. Z A+ Z B+ Z ACB = two rt. Zs.
Cor.— The exterior angle of a triangle is greater
than either of the interior and opposite angles.
46 THEORETICAL GEOMETRY BOOK I
56. — Exercises
y^l. Prove Theorem 10 by means of a st. line drawn
through the vertex || the base.
2. If two As have two Z s of one respectively equal to
two /s of the other, the third / of one is equal to the
third / of the other.
X,3. The sum of the Zs of a quadrilateral is equal to four
rt. Zs.
4. The sum of the zs of a pentagon is six rt. zs.
5. Each Z of a equilateral A is an z of 60°.
6. Find a point B in a given st. line CD such that, if
AB be drawn to B from a given point A, the Z ABC will
equal a given /.
7. Show that the bisectors of the two acute zs of a
rt.-zd A contain an Z of 135°.
T 8. If both pairs of opposite zs of a quadrilateral are
equal, the quadrilateral is a ||gm.
9. C is the middle point of the st. line AB. CD is drawn
in any direction and equal to CA or CB. Prove that ADB
is a rt. z«
. On AB, AC, sides of a A ABC, equilateral As ABD,
ACE are described externally. Show that DC = BE.
*j{ll. AB is any chord of a circle of which the centre is
O. AB is produced to C so that BC = BO. CO is joined,
cutting the circle at D and is produced to cut it again at
E. Show that Z AOE = three times Z BCD.
^12. If the exterior zs at B and C of a A ABC be
bisected and the bisectors be produced to meet at D, the
Z BDC equals half the sum of Zs ABC, ACB.
EXERCISES 47
13. Show that a A must have at least two acute Zs.
14. In an acute- zd A show that the _L from a vertex
to the opposite side cannot fall outside of the A.
1"). In an obtuse- zd A show that the _L from the vertex
of the obfcuse ^ on the opposite side falls within the A,
but that the J_ from the vertex of either acute / on the
opposite side falls outside of the A.
16. In a rt.-^d A where do the ±s from the vertices
on the opposite sides fall?
17. Only one JL can be drawn from a given point to a
given st. line.
18. Not more than two st. lines each equal to the same
given st. line can be drawn from a given point to a given
st. line.
19. D is a point taken within the A ABC. Join DB, DC;
and show, by producing BD to meet AC, that Z BDC > Z
BAG.
20. With compasses and ruler only, construct the follow-
ing ^s:-30°, 15°, 120°, 105°, 75°, 67J°, 150°, 195°, 210°,
240°, 255°, 285°, - 30°, - 753, - 135°.
21. If a transversal cut two st. lines so as to make the
interior Z_s on one side of the transversal together less
than two rt. Ls, the two lines when produced shall meet
on that side of the transversal.
22. The bisector of the exterior vertical L of an isosceles
A is II to the base.
23. Give a proof for the following method of drawing a
line through P J. AB : —
48
THEORETICAL GEOMETRY
BOOK I
First place the set-square in the position shown by the
dotted line, with its hypotenuse along AB,
Place a ruler along one of the sides of the set-square
and hold it firmly in that position.
Rotate the set-square through its right /-, thus bringing
the other side against the ruler, and slide the set-square
along the ruler to the position shown by the shaded A.
A line drawn through P, along the hypotenuse of the
set-square, is perpendicular to AB.
TRIANGLES 49
THEOREM 11
If one side of a triangle is greater than another
side, the angle opposite the greater side is greater
than the angle opposite the less side.
B C
Hypothesis. — ABC is a A having AB > AC.
To prove that Z ACB > Z ABC.
Construction. — From AB cut off AD = AC. Join DC.
Proof. — In A A DC,
AD = AC,
.-. Z ADC = L ACD. (1—3, p. 20.)
But L ACB> Z ACD,
.'. Z ACB> Z ADC.
In A BDC,
v BD is produced to A,
/. exterior Z ADC> interior and opposite
Z DBC. (I— 10, Cor., p. 45.)
But Z ACB> Z ADC;
much more .-. is Z ACB > Z ABC.
50 THEORETICAL GEOMETRY BOOK I
THEOREM 12
(Converse of Theorem 11)
If one angle of a triangle is greater than another
angle of the same triangle, the side opposite the
greater angle is greater than the side opposite the
less.
B C
Hypothesis. — In A ABC L B > Z C.
To show that AC>AB.
Proof. — If AC be not > AB,
then either AC = AB,
or AC<AB.
If AC = AB,
then Z B = Z C. (1—3, p. 20.)
But this is not so, .*. AC is not = AB.
If AC<AB,
then Z B < Z C. (I— 1 1, p. 49.)
But this also is not so, .*. AC is not < AB.
Hence v AC is neither = nor <AB,
.-. AC>AB.
57.— Exercises
1. The perpendicular is the shortest st. line
that can be drawn from a given point to a
given straight line.
The length of the JL from, a given point to a given st. line
is called the distance of the point from the line.
EXERCISES 51
X 2. A BCD is a quadrilateral, of which AD is the longest
side, and BC the shortest. Show that Z B > Z D, and that
Z C> Z A.
3. The hypotenuse of a rt.-z_d A is greater than either
of the other two sides.
X 4. A st. line drawn from the vertex of an isosceles A to
any point in the base is less than . either of the equal sides.
"* 5. A st. line drawn from the vertex of an isosceles A to
any point in the base produced is greater than either of
the equal sides.
6. If one side of a A be less than another, the L
opposite the less side is acute.
^ 7. If D be any point in the side BC of a A ABC, the
greater of the sides AB, AC, is greater than AD.
> 8. AB is drawn from A _L CD. E, F are two points in
CD on the same side of B, and sucli that BE<BF. Show
that AE < AF. Prove the same proposition when E, F are
on opposite sides of B.
> 9. ABC is a A having AB>AC. The bisector of Z A
meets BC at D. Show that BD > DC. Give a general
statement of this proposition.
10. ABC is a A having AB > AC. If the bisectors of
LS B, C meet at D, show that BD> DC.
11. Prove Theorem 11 from the
following construction : Bisect Z A
by AD which meets BC at D ; from
AB cut off AE = AC, and join
ED- .
12. The z_s at the ends of the greatest side of a A are
acute.
13. If AB>AD in the ||gm ABCD, Z ADB > Z BDC.
52 THEORETICAL GEOMETRY BOOK I
THEOREM 13
(Converse of Theorem 3)
If two angles of a triangle are equal to each
other, the sides opposite these equal angles are
equal to each other.
A
B c
Hypothesis. — In A ABC Z B = Z C.
To prove that AB = AC.
Proof.— If AB is not = AC,
let AB>AC.
Then Z C> z B. (I— 11, p. 49.)
But this is not so.
/. AB is not >AC.
Similarly it may be shown that
AB is not < AC.
. AB = AC.
58. — Exercises
1. An equiangular A is equilateral.
2. BD, CD bisect the L s ABC, ACB at the base of an
isosceles A ABC. Show that A DEC is isosceles.
. 3. ABC is a A having AB, AC produced to D, E re-
spectively. The exterior L s DBC, ECB are bisected by
EXERCISES 53
BF, CF, which meet at F. Show that, if FB = FC, the
A ABC is isosceles.
4. On the same side of AB the two As ACB, ADB have
AC = BD, AD = BC, and AD, BC meet at E. Show that
AE = BE.
5. On a given base construct a A having one of the L s
at the base equal to a given z_ , and the sum of the sides
equal to a given st. line.
6. On a given base construct a A having one of the L. s
at the base equal to a given L arid the difference of the
sides equal to a given st. line.
y 7. If the bisector of an exterior L of a A be || to the
opposite side, the A is isosceles.
y( 8. Through a point on the bisector of an L a line is
drawn || to one of the arms. Prove that the A thus formed
is isosceles.
9. A st. line drawn JL to BC, the base of an isosceles A
ABC, cuts AB at X and CA produced at Y. Show that
AXY is an isosceles A.
10. ACB is a rt.-^d A having the rt. L. at C. Through
X, the middle point of AC, XY is drawn || CB cutting AB
at Y. Show that Y is the middle point of AB.
11. The middle point of the hypotenuse of a rt.-Zd A is
equidistant from the three vertices.
12. The st. line joining the middle points of two sides of
a A is II to the third side.
13. Construct a rt.- L d A, having the hypotenuse equal to
one given st. line, and the sum of the other two sides
equal to another given st. line.
14. If one L of a A equals the sum of the other two,
show that the A is a rt.-z.d A.
54 THEORETICAL GEOMETRY BOOK I
THIRD CASE OF THE CONGRUENCE OF TRIANGLES
THEOREM 14
If two triangles have two angles and a side of
one respectively equal to two angles and the corre-
sponding side of the other, the triangles are
congruent
D,
B C E •
Hypothesis. — ABC, DEF are two As having Z A =
L D, L B = L E, and BC = EF.
To prove that A ABC = A DEF
Proof. — v L A = Z D,
and L B = L E,
But Z A+ Z B+ZC= Z D-f Z E-f- Z F. (I— 10, p. 45.)
/. Z C = Z F.
Apply A ABC to A DEF so that BC coincides with
the equal side EF.
v Z B = Z E,
.-. BA falls along ED, and A is on the line ED.
v Z C = Z F,
/. CA falls along FD, and A is on the line FD.
But D is the only point common to ED and FD,
/. A falls on D.
/. A ABC coincides with A DEF,
and . . A ABC = A DEF.
EXERCISES 55
59. — Exercises
C 1. If the bisector of an Z of a A be X to the opposite
side, the A is isosceles.
2. Any point in the bisector of an Z is equidistant from
the arms of the /.
3. In the base of a A find a point that is equidistant
from the two sides.
4. In a given st. line find a point that is equidistant
from two other given st. lines.
^ 5. Within a A find a point that is equally distant from
the three sides of the A.
6. Without a A find three points each of which is equally
distant from the three st. lines that form the A.
7. The ends of the base of an isosceles A are equidistant
from the opposite sides.
8. Two rt.-Zd As are congruent, if the hypotenuse and-
an acute / of one are respectively equal to the hypote-
nuse and an acute ^ of the other.
9. Construct a A with a side and two /s respectively
equal to a given st. line and two given /s.
10. The _L from the vertex of an isosceles A to the base,
bisects the base and the vertical /.
11. Prove I — 13 by drawing the bisector of the vertical /,
and using I — 14.
12. A ABC = A DEF and AX, DY are J_ to BC, EF
respectively. Prove that AX = DY.
13. A ABC = A DEF and AM, DN bisect Zs A, D and
meet BC, EF at M, N respectively. Prove that AM = DN.
14. If the diagonal AC of a quadrilateral A BCD bisects
the /s at A and C, AC is an axis of symmetry of ABCD.
15. The middle point of the base of an isosceles A is
equidistant from the equal sides.
56 THEORETICAL GEOMETRY BOOK I
THE AMBIGUOUS CASE IN THE COMPARISON OF
TRIANGLES
THEOREM 15
If two triangles have two sides of one respec-
tively equal to two sides of the other and have the
angles opposite one pair of equal sides equal to
each other, the angles opposite the other pair of
equal sides are either equal or supplementary.
FIG. 1
Hypothesis. — ABC, DEF are two As having AB = DE,
= DF and Z B = Z E.
To prove that either ZC = Z F,
or Z C + ^ F = two rt. Z s.
Proof. — Case I. Suppose z A = Z D. (Fig. 1.)
Then in the two As ABC, DEF,
v Z A = Z D,
and Z B = Z E,
But ZA+^B+ZC=ZD+ZE+ZF. (I— 10, p. 45.)
.'. Z C = Z F.
Case II. Suppose z A not = Z D. (Fig. 2.)
THE AMBIGUOUS CASE 57
Make Z EDG = Z BAG, and produce its arm to meet
EF, produced if necessary, at G.
!L A = L EDG,
L B = L E,
AB = DE,
J; G' } (I_14, p. 54)
and AC = DG. J
But DF = AC, (Hyp.)
:. DF = DG.
.. L DFG = L G. (1—3, p. 20.)
But L C - L G.
.. L C = L DFG.
Z. DFG + L DFE = two rt. /s,
/. Z_ C 4- L DFE = two rt. Zs.
NOTE. — There are six parts in a triangle, viz., three
sides and three angles, and in the cases in which the
congruence of two triangles has been established three
parts of one triangle, one at least a side, have been given
respectively equal to the corresponding parts of the
other.
The following general cases occur: —
1. Two sides and the contained angle. The triangles
are congruent— Theorem 2.
2. Three sides. The triangles are congruent —
Theorem 4.
3. Two angles and a side. The triangles are con-
gruent— Theorem 14.
4. Two sides and an angle opposite one of them.
In this case the triangles are congruent if the angle is
opposite the greater of the two sides — §60, Ex. 3, but
58 THEORETICAL GEOMETRY BOOK I
if the angle is opposite the less of the two sides, they
are not necessarily congruent — Theorem 15.
5. Three angles. The triangles are not necessarily
congruent — §60, Ex. 7.
60.— Exercises
1. If two rt.-Zd As have the hypotenuse and a side of
one respectively equal to the hypotenuse and a side of the
other, the As are congruent.
2. If the bisector of the vertical Z of a A also bisects
the base, the A is isosceles.
3. If two As have two sides of one respectively equal
to two sides of the other and the Zs opposite the greater
pair of equal sides equal to each other, the As are
congruent.
4. Construct a A having given two sides and the Z
opposite one of them.
When will there be : (a) no solution, (ft) two solutions,
(c) only one solution?
5. If two Zs of a A be bisected and the bisectors be
produced to meet, the line joining the point of intersection
to the vertex of the third Z bisects that third Z. Hence.—
The bisectors of the three Zs of a A pass through one
point.
6. If two exterior Zs of a A be bisected and the
bisectors be produced to meet, the line joining the point
of intersection of the bisectors to the vertex of the third
Z of the A bisects that third Z.
7. Draw diagrams to show that if the three Ls of one
A are respectively equal to the three i_ s of another A, the
two As are not necessarily congruent.
INEQUALITIES 59
INEQUALITIES
THEOREM 16
Any two sides of a triangle are together greater
than the third side.
\A
B DC
Hypothesis. — ABC is a A.
To prove that AB + AC > BC.
Construction. — Bisect z A and let the bisector meet
BC at D.
Proof. — ADC is an exterior L of A ABD,
/. Z ADC> / BAD. (I— 10, Cor., p. 45.)
But Z BAD = Z DAC.
.'. Z ADC> / DAC.
AC> DC. (I— 12, p. 50.)
Similarly it may be shown that
AB> BD.
.. AB+AOBD-f DC,
i.e., AB + AOBC.
In the same manner it may be shown that
AB + BC> AC and that AC + CB > AB.
60 THEORETICAL GEOMETRY BOOK I
Cor. — The difference between any two sides of a
triangle is less than the third side.
B
ABC is a A.
It is required to show that AB - AC < BC.
AB < AC + BC. (1—16, p. 59.)
From each of these unequals take AC,
and AB - AC < BC.
In the same manner it may be shown that AB - BC
< AC and that BC - AC < AB.
61.— Exercises
1. Show that the sum of any three sides of a quadri-
lateral is greater than the fourth side.
2. The sum of the four sides of a quadrilateral is greater
than the sum of its diagonals.
3. The sum of the diagonals of a quadrilateral is greater
than the sum of either pair of opposite sides.
4. The sum of the st. lines joining any point, except the
intersection of the diagonals, to the four vertices of a
quadrilateral, is greater than the sum of the diagonals.
5. If any point within a A be joined to the ends of a side
of the A, the sum of the joining lines is less than the sum
of the other two sides of the A.
EXERCISES 61
6. If any point within a A be joined to the three vertices
of the A, the sum of the three joining lines is less than
the perimeter of the A, but greater than half the perimeter.
7. The sum of any two sides of a A is greater than twice
the median drawn to the third side.
8. The median of a A divides the vertical /_ into parts, of
which the greater is adjacent to the less side.
9. The perimeter of a A is greater than the sum of the
three medians.
10. A and B are two fixed points, and CD is a fixed st.
line. Find the point P in CD, such that PA -h PB is the
least possible:
(a) When A and B are on opposite sides of CD;
(b} When A and B are on the same side of CD.
11. A and B are two fixed points, and CD is a fixed st.
line. Find the point P in CD, such that the difference
between PA and PB is the least possible;
(a) When A and B are on the same side of CD;
(b) When A and B are on opposite sides of CD.
12. Prove Theorem 16 by producing BA to E, making
AE = AC, and joining EC.
13. Prove that the shortest line which can be drawn
with its ends on the circumferences of two concentric
circles, will, when produced, pass through the centre.
14. Prove the Corollary under Theorem Ifi, (a) by cutting
off from AB a part AD = AC and joining DC; (b) by pro-
ducing AC to E making AE = AB and joining BE.
62 THEORETICAL GEOMETRY BOOK I
THEOREM 17
If two triangles have two sides of one respec-
tively equal to two sides of the other but the
contained angle in one greater than the contained
angle in the other, the base of the triangle which
has the greater angle is greater than the base of
the other.
Hypothesis. — ABC, DEF are two As having AB = DE,
AC = DF and Z BAG > Z EDF.
To show that BC> EF.
Construction. — Make z EDG = Z BAG and cut off
DG = AC, or DF. Join EG. Bisect Z FDG and let the
bisector meet EG at H. Join FH.
Proof.—
' AB = DE,
In AS ABC, DEG, \ AC = DG,
Z A = Z EDG,
BC = EG. (1—2, p. 16.)
DF = DG,
In As FDH, GDH, \ DH is common,
Z FDH = Z GDH,
.-. FH = HG.
In A EHF, EH + HF > EF. (1—16, p. 59.)
INEQUALITIES 63
But HF = HG,
EH + HG > EF.
i.e., EG > EF.
But BC = EG,
/. BC > EF.
64 THEORETICAL GEOMETRY BOOK I
THEOREM 18
(Converse of Theorem 17)
If two triangles have two sides of one respec-
tively equal to two sides of the other but the base
of one greater than the base of the other, the
triangle which has the greater base has the greater
vertical angle.
Hypothesis. — ABC, DEF are two As having AB = DE,
AC = DF and BC > EF.
To prove that z A > Z D.
Proof.— If Z A is not > Z D,
either z A = Z D,
or z A < Z D.
(1) If Z A = Z D.
In As ABC, DEF,
z A = Z D,
(1—2, p. 16.)
But this is not so.
.-. Z A is not = Z D.
(2) If Z A < Z D.
INEQUALITIES— EXERCISES 65
In AS ABC, DEF,
(I— 17, p. 62.)
But this is not so.
/. Z A is not < Z D.
Then since Z A is neither = nor < Z D,
/. Z A > Z D.
62. — Exercises
/ 1. ABCD is a quadrilateral having AB = CD and Z BAD>
Z ADC. Show that Z BCD > Z ABC.
2. In A ABC, AB > AC and D is the middle point of
BC. If any point P in the median AD be joined to B
and C, BP > CP.
If AD be produced to any point Q show that BQ < QC.
/3. D is a point in the side AB of the A ABC. AC is
produced to E making CE = BD. BE and CD are joined.
Show that BE>CD.
4. If two chords of a circle be unequal the greater sub-
tends the greater angle at the centre.
5. Two circles have a common centre at O. A, B are
two points on the inner circumference and C, D two on
the outer. L AOC > Z BOD. Show that AC > BD.
4^6. CD bisects AB at rt. Zs. A point E is taken not in
CD. Prove that EA, EB are unequal.
/ 7. In A ABC, AB > AC. Equal distances BD, CE are cut
off from BA, CA respectively. Prove BE > CD.
y 8. In A ABC, AB > AC. AB, AC are produced to D,
E making BD = CE. Prove CD > BE.
66 THEORETICAL GEOMETRY BOOK I
PARALLELOGRAMS
THEOREM 19
Straight lines which join the ends of two equal
and parallel straight lines towards the same parts
are themselves equal and parallel.
Hypothesis. — AB, CD are = and ||.
To prove that (1) AC = BD,
(2) AC || BD.
Construction. — Join AD.
Proof.— . AB || CD,
and AD is a transversal,
.'. Z BAD = Z CDA. (1—8, p. 40.)
IBA = CD,
AD is common,
Z BAD = Z CDA,
.. BD = AC,
and Z BDA = Z CAD,
".* transversal AD
makes Z BDA = Z CAD,
. . BD || AC. (1—6, p. 36.)
^
J a-
PARALLELOGRAMS 67
THEOREM 20
In any parallelogram:
(1) The opposite sides are equal ;
(2) The opposite angles are equal;
(3) The diagonal bisects the area;
(4) The diagonals bisect each other.
B
Hypothesis. — ABCD is a ||gm, AC, BD its diagonals.
To prove that (1) AD = BC and AB = CD.
(2) Z BAD = Z BCD and Z ABC = Z ADC.
(3) A ABC = AACD.
(4) AE = EC and BE - ED.
Proof. — v AC cuts || lines AD, BC,
/. Z DAC = Z ACB. (1—8, p. 40.)
v AC cuts (I lines DC, AB,
/. Z DCA = Z CAB.
{Z DAC = Z ACB,
Z DCA = Z CAB,
AC is common,
.'. (1) AD = BC, and CD = AB, \
(2) also Z ADC - Z ABC, }(I— 14,p.54.;
(3) and A ADC = A ABC. J
Similarly it may be shown that Z BAD = Z BCD.
AD = BC,
Z DAE = Z BCE,
Z ADE = Z CBE,
(1-14, p. 54.)
and DE = EB.
68
THEORETICAL GEOMETRY
BOOK I
63. Definitions- — A parallelogram of which the angles
are right angles is called a rectangle.
A rectangle of which all the sides are equal to each!*
other is called a square.
A figure bounded by more than four straight lines is
called a polygon.
The name polygon is sometimes used for a figure
having any number of sides.
A polygon in which all the sides are equal to each
other and all the angles are equal to each other is called
a regular polygon.
64.— Exercises
1. The diagonals of a rectangle are equal to each other.
2. If the diagonals of a ||gm are equal to each other, the
||gm is a rectangle.
3. A rectangle has two axes of symmetry.
4. A square has four axes of symmetry.
>* 5. The st. line joining the
middle points of the sides of a
A is II the base, and equal to*
half of it.
NOTE. — D, E are the middle
points of AB, AC. Produce DE
to F 'making EF = DE. Join
FC.
"1* 6. Of two medians of a A
each cuts the other at the point
of trisection remote from the
vertex.
NOTE. — Medians BE, CF cut
at G. Bisect BG, CG at H, K.
B~~ ~~C Join FH, HK, KE, EF.
7. The medians of a A pass through one point.
EXERCISES 69
Definition. — The point where the medians of a A inter-
sect is called the centroid of the A.
x 8. A st. line drawn through the middle point of one side
of a A, I! to a second side, bisects the third side.
9. In any ||gm the diagonal which joins the vertices of the
obtuse zs is shorter than the other diagonal.
X 10. If two sides of a quadrilateral be ||, and the other
two be equal to each other but not ||, the diagonals of the
quadrilateral are equal.
11. Through a given point draw a st. line, such that
the part of it intercepted between two given || st. lines is
equal to a given st. line.
Show that, in general, two such lines can be drawn.
12. Through a given point draw a st. line that shall be
equidistant from two other given points.
Show that, in general, two such lines can be drawn.
X 13. Draw a st. line || to a given st. line, and such that
the part of it intercepted between two given intersecting
lines is equal to a^^^^^^^^
7<:14. BAG is a given /_, and P is a given point. Draw a
st. line terminated in the st. lines AB, AC and bisected
at P.
* 15. Construct a A having given the middle points of
the three sides.
16. If the diagonals of a ||gm cut each other at rt. z_s,
the || gm is a rhombus. *•*
X'17. Every st. line drawn through the intersection of the
diagonals of a ||gm, and terminated by a pair of opposite
sides, is bisected, and bisects the ||gm.
18. Bisect a given ||gm by a st. line drawn through a
given point.
19. Divide a given A into four congruent As.
70
THEORETICAL GEOMETKY
BOOK I
)( 20. The bisectors of two opposite zs of a ||gm are || to
each other.
/21. In the quadrilateral ABCD, AB || CD and AD = BC.
Prove that (1) Z C = Z D ; (2) if E, F are the middle
points of AB, CD respectively, EF J_ AB.
22. On a given st. line construct a square.
23. Construct a square having its diagonal equal to a
given st. line.
K 24. ABC is a A and DE a st. line. Draw a st. line
= DE, || BC and terminated in AB, AC, or in these lines
produced.
V'25. Inscribe a rhombus in a given ||gm, such that one vertex
of the rhombus is at a given point in a side of the ||gm.
26. ABC is an isosceles A in which AB = AC. From P,
any point in BC, PX, PY are drawn _J_ AB, AC respectively
and BM is J_ AC. Prove that PX + PY = BM.
If P is taken on CB produced, prove that PY - PX =
BM.
27. TtWThfddle point of the
hypotenuse of a rt.--^d A is
equidistant from the three
vertices.
NOTE. — Through D, the mid-
dle point of the hypotenuse AB,
draw DE || BC. Join DC.
28. ABCD is a quadrilateral
in which AB II CD. E, F, G, H
are the middle points of BC,
BD, AC, AD. Prove that: (1)
the st. line through E || AB, or
DC, passes through F, G and
H; (2) HE = half the sum of
ABand DC; (3) GF = half the difference of AB and DC.
B
EXERCISES
71
29. E, F, G, H are the middle points of the sides AB, BC,
CD, DA of the quadrilateral
ABCD. Prove that EFGH is a
|| gm. Show also that: (1) the
perimeter of EFGH = AC + BD;
(2) if AC = BD, EFGH is a
rhombus ; (3) if AC _L BD, EFGH
is a rectangle ; (4) if AC = and
J_ BD, EFGH is a square.
30. The middle points of a
pair of opposite sides of a
quadrilateral and the middle points of the diagonals are
the vertices of a ||gm.
31. The st. lines joining the middle points of the opposite
sides of a quadrilateral and the st. line joining the middle
points of the diagonals are concurrent.
A
72
THEORETICAL GEOMETRY
THEOREM 21
BOOK I
The sum of the interior angles of a polygon of n
sides is (2/7-4) right angles.
Hypothesis. — ABODE, etc., is a closed polygon of n
sides.
To prove that the sum of the interior angles is
(2n — 4) rt. Ls.
Construction. — Take any point P within the polygon
and join P to the vertices.
Proof. — The polygon is divided into n As PAB,
PBC, PCD, etc.
The sum of the interior z_s of each A is two
rt. LS. (I— 10, p. 45.)
.-. the sum of the LS of the n As is 2n rt. ^s.
But the ^LS of the n As make up the interior Ls of
the polygon together with the z_s about the point P.
And the sum of the Ls about P equals 4 rt. z_s.
/. the sum of the interior ^.s of the polygon =
(2n - 4) rt. LS.
EXERCISES 73
Cor.— If the sides of a polygon are produced in
order, the sum of the exterior angles thus formed
is four right angles.
If the polygon has n sides, the sum of all the st.
Ls at the vertices = 2n rt. ^.s.
But, the sum of the interior z_s = (2n — 4) rt.
LS. (1—21, p. 72.)
.'. , subtracting, L a + L b + etc. = 4 rt. z_s.
65.— Exercises
1. Find the number of degrees in an exterior L. of an
equiangular polygon of twelve sides.
Hence, find the number of degrees in each interior L .
2. Find the number of degrees in each L of (a) an equi-
angular pentagon; (b) an equiangular hexagon; (c) an
equiangular octagon; (d) an equiangular decagon.
3. Each L of an equiangular polygon contains 162°.
Find the number of sides.
4. Each z_ of an equiangular polygon contains 170°.
Find the number of sides.
5. Show that the space around a point may be exactly
filled in by six equilateral As, four squares, or three equi-
angular hexagons. Draw the diagram in each case.
74 THEORETICAL GEOMETRY BOOK I
CONSTRUCTION
PROBLEM 8
To divide a straight line into any number of equal
parts.
A F Q R 3 B
Let AB be the given st. line.
To divide AB into five equal parts.
Construction. — From A draw a st. line AC.
From AC cut off five equal parts AD, DE, EF, FG, GH.
Join HB.
Through D, E, F, G draw lines || HB cutting AB at
P, Q, R, S.
AB is divided into five equal parts at P, Q, R, S.
Proof.— Through D, E, F, G draw DK, EL, FM, GN || AB.
V AE cuts the parallels AP, DK,
/. L EDK = L DAP. (1—9, p. 42.)
V AE cuts the parallels DP, EQ,
/. L ADP = L DEQ.
IL DAP = L EDK,
L ADP = L DEK,
AD = DE,
AP = DK (1—14, p. 54.)
But PQ = DK. (1—20, p. 67.)
PQ = AP.
LOCI
75
Similarly it may be shown that each of QR, RS,
SB = AP.
By this method a st. line may be divided into any
number of equal parts.
Loci
66. Example I.— A is a point and from A straight
lines are drawn in different directions in the same
plane.
On each line a distance of one inch is measured
from A and the resulting points are B, C, D, etc.
Is there any one line that contains all of the points
in the plane that are at a distance of one inch
from A?
To answer this question describe a circle with
centre A and radius one inch. The circumference of
this circle is a line that passes through all the points.
76 THEORETICAL GEOMETRY BOOK I
Mark any other point P on the circumference.
What is the distance of P from A? From the defini-
tion of a circle the answer to this question is one
inch.
If any point Q be taken within the circle, its
distance from A is less than one inch, and if any
point R be taken without the circle, its distance from
A is greater than one inch.
Thus every point in the circumference satisfies the
condition of being just one inch from A, and no point,
in the plane, that is not on the circumference does
satisfy this condition.
This circumference is called the locus of all points
in the plane that are at a distance of one inch
from A.
Example 2: — AB is a straight line of indefinite
length, to which any number of perpendiculars are
drawn.
jc ;D IE IF
1 ~
A i
i 1 ,
! :
i I
*Qi
i
•
1
3f i
ei r0
H 3^ |K
On each of these perpendiculars a distance of one
centimetre is measured from AB, and the resulting
points are C, D, E, etc.
Are there any lines that contain all of the points,
such as C, D, etc., that are at a distance of one centi-
metre from AB?
LOCI 77
Draw two straight lines parallel to AB, each at a
distance of one centimetre from AB, and one or other
of these lines will pass through each of the points.
Any point P in CF, or in GK, is at a distance of one
centimetre from AB ; any point Q in the space between
CF and GK is less than one centimetre from AB, and
any point R in the plane and neither between CF and
GK nor in one of these lines is more than one centi-
metre from AB.
Thus every point in CF and GK satisfies the con-
dition of being just one centimetre from AB, and no
point outside of these lines and in the plane does
satisfy this condition.
The two lines GF, GK make up the locus of all
points in the plane that are at a distance of one
centimetre from AB.
Definition. — When a figure consisting of a line
or lines contains all the points that satisfy a given
condition, and no others, this figure is called the
locus of these points.
67. In place of speaking of the "locus of the points
which satisfy a given condition," the alternative ex-
pression "locus of the point which satisfies a given
condition" may be used.
Suppose a point to move in a plane so that it
traces out a continuous line, but its distance from a
fixed point A in the plane is always one inch ; then it
must move on the circumference of the circle of centre
A and radius one inch, and the locus of the point in
its different positions is that circumference.
78 THEORETICAL GEOMETRY BOOK 1
The following definition of a locus may thus be
given as an alternative to that in § 66.
Definition.— If a point moves on a line, or on lines,
so that it constantly satisfies a given condition, the
figure consisting of the line, or lines, is the locus
of the point.
THEOREM 22
The locus of a point which is equidistant from two
given points is the right bisector of the straight
line joining the two given points.
B
Hypothesis. — p is a point equidistant from A and B.
To prove that P is on the right bisector of AB,
Construction. — Bisect AB at C.
Join PC, PA, PB.
Proof.—
( PA = PB,
'
In AS PAC, PBC, ' AC = CB,
(^ PC is common,
/. A PAC = A PCB, (1—4, p. 22.)
-'. L PCA = L PCB,
and /. P is on the right bisector of AB.
LOCI
THEOKEM 23
The locus of a point which is equidistant from two
given intersecting straight lines is the pair of
straight lines which bisect the angles between the
two straight lines.
Hypothesis. — AB, CD are two st. lines cutting at
E; GF, HK are the bisectors of Ls made by AB, CD.
To prove that the locus of a point equidistant from
AB, and CD consists of GF and HK.
Construction. — Take any point P in GF.
PX J_ AB, PY J_ CD.
Draw
Proof.—
In As PEX, PEY,
Z PEX = Z PEY,
Z PXE = Z PYE,
PE is common,
PX = PY.
(1-14, p. 54.)
.-. every point in GF is equidistant from AB and CD.
Similarly it may be shown that every point in
HK is equidistant from AB and CD.
/. the locus of points equidistant from AB, CD consists
of GF and HK.
80 THEORETICAL GEOMETRY BOOK I
68. Problem: -To find the point that is equally
distant from three given points, that are not in the
same straight line.
Let A, B, C be the three given points.
It is required to find a point equally distant from
A, B and C.
Draw EF the locus of all points that are equally
distant from A and B. (I — 22, p. 78.)
,F
*C
Draw GH the locus of all points that are equally
distant from B and C.
Let EF and GH meet at K.
Then K is the required point.
K is on EF, .'. KA = KB.
K is on GH, .. KB = KG.
Consequently K is equally distant from A, B and C.
69.— Exercises
1. Find the locus of the centres of all circles that pass
through two given points.
2. Describe a circle to pass through two given points
and have its centre in a given st. line.
3. Describe a circle to pass through two given points
and have its radius equal to a given st. line. Show that
EXERCISES 81
generally two such circles may be described. When will
there be only one? and when none?
4. Find the locus of a point which is equidistant from
two given || st. lines.
5. In a given st. line find two points each of which is
equally distant from two given intersecting st. lines.
When will there be only one solution?
6. Find the locus of the vertices of all As on a given
base which have the medians drawn to the base equal to
a given st. line.
7. Find the locus of the vertices of all As on a given
base which have one side equal to a given st. line.
8. Construct a A having given the base, the median
drawn to the base, and the length of one side.
9. Find the locus of the vertices of all As on a given
base which have a given altitude.
10. Construct a A having given the base, the median
drawn to the base, and the altitude.
11. Construct a A having given the base, the altitude
and one side.
12. Find the locus of a point such that the sum of its
distances from two given intersecting st. lines is equal to
a given st. line.
13. Find the locus of a point such that the difference of
its distances from two given intersecting st. lines is equal
to a given st. line.
14. Find the locus of the vertices of all As on a given
base which have the median drawn from one end of the
base equal to a given st. line.
82 THEORETICAL GEOMETRY BOOK I
15. Show that, if the ends of a st. line of constant
length slide along two st. lines at rt. L s to each other,
the locus of its middle point is a circle.
Ifi. AB is a st. line and C is a point at a distance of 2
cm. from AB. Find a poirit which is 1 cm. from AB and
4 cm. from C. How many such points can be found?
17. Two st. lines, AB, CD, intersect each other at an L
of 45°. Find all the points that are 3 cm. from AB and
2 cm. from CD.
18. ABC is a scalene A. Find a point equidistant from
AB and AC, and also equidistant from B and C.
19. Find a point equidistant from the three vertices of a
given A.
20. Find four points each of which is equidistant from
the three sides of a A.
NOTE. — Produce each side in both directions.
21. Find the locus of a point at which two equal seg-
ments of a st. line subtend equal L s.
2 '2. Find the locus of the centre of a circle which shall
pass through a given point and have its radius equal to a
given st. line.
23. A st. line of constant length remains always || to
itself, while one of its extremities describes the circum-
ference of a fixed circle. Find the locus of the other
extremity.
24. The locus of the middle points of all st. lines drawn
from a fixed point to the circumference of a fixed circle
is a circle.
MISCELLANEOUS EXERCISES 83
Miscellaneous Exercises
1. If a st. line be terminated by two ||s, all st. lines
drawn through its middle point and terminated by the
same ||s are bisected at that point.
2. If two lines intersecting at A
be respectively II to two lines inter-
secting at B, each L at A is either
equal to or supplementary to each
L at B.
\
3. If two lines intersecting at
^ A be respectively J_ to two lines
B\ " intersecting at B, each L at A is
either equal to or supplementary
L at B>
^
\ "
\
\
4. If from any point in the
bisector of an L st. lines be drawn
|| to the arms of the L and
terminated by the arms, these st. lines are equal to each
other.
\ 5. In the base of a 2^ find a point such that the st.
lines drawn from that point || to the sides of the A and
terminated by the sides are equal to each other.
6. One L of an isosceles A is half each of the others.
Calculate the L s.
7. If the ± from the vertex of a A to the base falls
within the A, the segment of the base adjacent to the
greater side of the A is the greater.
^S. If a star-shaped figure be formed by producing the
alternate sides of a polygon of n sides, the sum of the /s
at the points of the star is (2 n - 8) rt. L s.
V 9. In a quadrilateral ABCD, Z A = Z B and Z C = Z D.
Prove that AD = BC.
84 THEORETICAL GEOMETRY BOOK 1
10. The bisectors of the L s of a ||gm form a rectangle,
the diagonals of which are || to the sides of the original
1 1 gin ; and equal to the difference between them.
11. From A, B the ends of a st. line ±s AC, BD are
drawn to any st. line. E is the middle point of AB.
Show that EC = ED.
X12. If through a point within a A three st. lines be
drawn from the vertices to the opposite sides, the sum of
these st. lines is greater than half the perimeter of the A.
13. A, D are the centres of two circles, and AB, DE are
two || radii. EB cuts the circumferences again at C, F.
Show that AC || DF.
X 14. The bisectors of the interior z_ s of a quadrilateral
m a quadrilateral of which the opposite L s are supple-
mentary.
15. In a given square inscribe anJequilateral A having
one vertex at a vertex of the
A
ibe anJ
squai^
16. Through two given points draw two st. lines, forming
an equilateral A with a given st. line.
17. Draw an isosceles A having its base in a given st.
line, its altitude equal to a given st. line, and its equal
sides passing through two given points.
18. If a J_ be drawn from one end of the base of an
isosceles A to the opposite side, the L between the ± and
the base = half the vertical L of the A.
^19. If any point P in AD the bisector of the Z A of A
ABC be joined to B and C, the difference between PB aJnd
PC is less than the difference between AB and AC.
X20. If any point P in the bisector of the exterior i. at
A in the A ABC be joined to B and C, PB -f PC > AB +
AC.
MISCELLANEOUS EXERCISES 85
21. BAC is a rt. L. and D is any point. DE is drawn _|_
AB and produced to F, making EF = DE. DG is drawn J_
AC and produced to H, making GH = DG. Show that
F, A, H are in the same st. line.
22. Construct a A having its perimeter equal to a given\
st. line and its L s respectively equal to the L s of a/
iven A. *
23. In any quadrilateral, the sum of the exterior L. s at
one pair of opposite vertices = the sum of the interior _ s
at the other vertices.
24. If the arms of one z_ be respectively |] to the arms
of another L , the bisectors of the ^s are either II or J_.
In a given A inscribe a ||gm the diagonals of which
intersect at a given point.
26. Show that the _|_s from the centre of a circle to two
equal chords are equal to each other.
27. Construct a quadrilateral having its sides equal to
four given st. lines and one L equal to a given L .
The bisector of Z A of A ABC meets BC at D and!
produced to E. Show that Z ABC + Z ACE = twice 7
-3
vw The bisectors of L s A and B of A ABC intersect at D.
Show that Z ADB = 90° + half of Z C.
V@ The sides AB, AC of a A ABC are bisected at D, E;
and BE, CD are produced to F, G, so that EF = BE and
DG = CD. Show that F, A, G are in the same st. line, and
that FA = AG.
(SD ABC is an isosceles A, having AB = AC. AE, AD
are equal parts cut off from AB, AC respectively. BD, CE
cut at F. Show that FBC and FDE are isosceles As.
86 THEORETICAL GEOMETRY BOOK I
32. In a A ABC, the bisector of Z A and the right
bisector of BC meet at D. DE, DF are drawn ± AB, AC
respectively. Show that the point D is not within the A,
that AE = AF and that BE = CF.
Y 33. A BCD is a quadrilateral having Z B = Z C and
AB < CD. Prove that Z A > Z D.
34. Through a given point draw a st. line cutting two
intersecting st. lines and forming an isosceles A with them.
Show that two such lines can be drawn through the
given point.
35. If ACB be a st. line and ACD, BCD two adjacent
Zs, any || to AB will meet the bisectors of these zs in
points equally distant from where it meets CD.
m.
36. Inscribe a square in a given
\\ equilateral A.
NOTE. — Draw a sketch as in the
diagram given here. Join AE.
What is the number of degrees in
B D EC z CAE ?
37. ABC is a A, AX is _L BC, and AD bisects Z BAG.
Show that L XAD equals half the difference of Z s B and C.
38. Construct a ||gm having its diagonals and a side
respectively equal to three given st. lines.
39. Find a point in each of two || st. lines such that the
two points are equally distant from a given point and the
st. line joining them subtends a rt. z at the given point.
40. P, Q are two given points on the same side of a
given st. line BC. Find the position of a point A in BC
such that L PAB = L QAC.
NOTE. — If P, Q are two points on a billiard table and
BC the side of the table, a ball starting from P and reflected
from BC at A would pass through Q.
MISCELLANEOUS EXERCISES 87
41. Find the path of a billiard ball which, starting from
a given point, is reflected from the four sides of the table
and passes through another given point.
42. BAG is a given l_ and D, E are two given st. lines.
Find a point P such that its distances from AB, AC equal
D, E respectively.
43. Find in a side of a A a point such that the sum of
the two st. lines drawn from the point || to the other sides
and terminated by them is equal to a given st. line.
44. AEB, CED are two st. lines, and each of the quadri-
laterals CEAF, BEDG is a rhombus. Prove that PEG is a
st. line.
45. F is a point within the A ABC such that L FBC =
L FCB. BF, CF produced meet AC, AB at D, E respec-
tively. Prove that if L AFD = L AFE, A ABC is isosceles.
46. D is a point in the base BC of an equilateral
A ABC. E is the middle point of AD. Prove that
EC > ED.
47. ABC is a A of which L BAC is obtuse, O a point
within it; BO, CO meet AC, AB at D, E respectively.
Prove that BD + CE > BE + ED + DC.
48. D, E, F are points in the sides BC, CA, AB of an
equilateral A and are such that BD =CE=AF. If
AD, BE, CF do not all pass through one point, they form
an equilateral A.
49. The bisector of Z A of A ABC meets BC at D.
DE, DF drawn || AB, AC respectively meet AC, AB at E, F.
Prove that AEDF is a rhombus.
50. Through each angular point of a A a st. line is
drawn || the opposite side : prove that the A formed by
these three st. lines is equiangular to the given A.
88 THEORETICAL GEOMETRY BOOK I
51. AD, BE, CF respectively bisect the interior Z A and
the exterior zs at B and C of the A ABC. Show that
no two of the lines AD, BE, CF can be ||.
52. DE is || to the base AB of the isosceles A CAB and
cuts CA, CB, or those sides produced, at D, E respectively.
AE, BD cut at F. Prove that DEF is an isosceles A.
53. Through A, B the extremities of a diameter of a
circle || chords AC, BD are drawn. Prove that AC = BD ;
and that CD is a diameter of the circle.
54. The median drawn from the vertex of a A is >, =
or < half the base according as the vertical z is acute,
right or obtuse.
55. ABC is a A, obtuse- zd at C; st. lines are drawn
bisecting CA, CB at rt. zs, cutting AB in D, E respec-
tively. Prove that Z DCE is equal to twice the excess of
Z ACB over a rt. z.
56. "With one extremity C of the base BC of an isosceles
A ABC as centre, and radius CB, a circle is described
cutting AB, AC at D, E respectively. Prove that DE || to
the bisector of Z B.
57. In A ABC side BC is produced to D. Prove that
the Z between the bisectors of zs ABC, ACD = half the
Z A.
58. Through the vertices of A ABC, st. lines falling
within the A are drawn making equal zs BAL, CBM,
ACN ; if these lines intersect in D, E, F, prove A DEF
equiangular to A ABC.
59. If the i_ between two adjacent sides of a ||gm be in-
creased, while their lengths do not alter, the diagonal through
the point of intersection will decrease.
MISCELLANEOUS EXERCISES 89
60. A, B, C are three given points. Find a point equi-
distant from A, B and such that its distance from C equals
a given st. line. When is the problem impossible 1
91. Through a fixed point draw a st. line which shall
make with a given st. line adjacent /s the difference of
which = a given z_.
62. Construct a A having given one L and the lengths of
the J_s from the vertices of the other z_s on the opposite
sides.
63. Construct an isosceles A having given the vertical i_
and the altitude.
64. Construct an isosceles A having given the perimeter
and altitude.
65. Prove that the quadrilateral formed by joining the
extremities of two diameters of a circle is a rectangle.
66. In a given ||gm inscribe a rhombus, such that one
diagonal passes through a given point.
67. St. lines are drawn from a given point to a given
st. line. Find the locus of the middle points of the st.
lines.
68. St. lines are drawn from a given point to the cir-
cumference of a given circle. Find the locus of the middle
points of the st. lines.
69. The sum of the J_s from any point within an equi-
lateral A to the three sides is equal to the altitude of
the A.
70. Draw a square which has the sum of a side and a
diagonal equal to 3 inches.
71. Draw a square in which the difference between a
diagonal and a side is 1 inch.
90 THEORETICAL GEOMETRY BOOK I
72. Draw a rectangle having one side 2 inches in length,
and subtending an z_ of 40° at the point of intersection
of the diagonals.
(Use a protractor in Exercises 72 to 86.)
73. Draw a ||gm with diagonals 2 inches and 4 inches
and their L of intersection 50°.
74. Draw a ||gm with diagonals 4 inches and 7 inches
and one side 5 inches.
75. Draw a ||gm with side 3 inches, diagonal 2J inches
and / 35°. Show that there are two solutions.
76. Draw a ||gm with side 2f inches, L 70° and
diagonal opposite L of 70° equal to 4 inches.
77. Draw a rectangle having the perimeter 8 inches and
an L between the diagonals 80°.
78. Draw a rectangle having the difference of two sides
1 inch and an i_ between the diagonals 50°.
79. Draw a rectangle which has the perimeter 9 inches
and a diagonal 3J inches.
80. Draw an L of 55°. Find within the /_ a point
which is 1 inch from one arm and 2 inches from the
other.
81. Construct a A in which side a - 7 cm., b + c =
10-6 cm. and L A = 78°.
82. Construct a A with perimeter 4 inches and /_s 70°
and 50°.
83. AB, CD are two || st. lines; P, Q two fixed points.
Find a point equidistant from AB, CD and also equidistant
from P and Q. When is this impossible?
MISCELLANEOUS EXERCISES 91
84. Through two given points on the same side of a
given st. line draw two st. lines so as to form with the
given st. line an equilateral A.
85. Construct a rhombus with one diagonal 2 inches and
the opposite i_ 100°.
80. Construct a A in which a = 8 cm., b — c = 2 cm.,
L C = £0°.
87. Squares ABGE, ACHF are described externally on
two sides of a A ABC Prove that the median AD of the
A is _L EF and equal to half of EF.
NOTE. — Rotate A ABC through a rt. L making AC
coincide with AF.
88. Prove also in Ex. 87 that EC is _L and = BF.
89. Trisect a rt. /_.
• 90. From any point in the base of an isosceles A st.
lines are drawn || to the equal sides and terminated by
them. Prove that the sum of these lines = one of the
equal sides.
91. ABC is a st. line such that AB = BC. J_s are
drawn from A, B, C to another st. line EF. Prove that
the J_ from B = half the sum of the JLs from A and C,
unless EF passes between A and C, and then the _L from
B = half the difference of the J_s from A and C.
92. AD is the bisector of L A of A ABC, and M the
middle point of BC. BE and CF are J_ AD. Prove that
ME = MF.
93. E, F are the middle points of AD, BC respectively
in the ||gm ABCD. Prove that BE, DF trisect AC.
94. Find a point P in the side AC of a A ABC so that
AP may be equal to the _L from P to BC.
92 THEORETICAL GEOMETRY BOOK I
95. If the st. line AB be bisected at C and produced to
D, prove that CD is half the sum of AD, BD.
96. In A ABC side AC > side AB ; AX j_ BC and AD is
a median. Prove that (1) Z CAX > Z BAX; (2) Z CAD <
Z DAB; (3) the bisector of Z BAC falls between AX and
AD.
97. The median of a A ABC drawn from A is not less
than the bisector of Z A.
98. In a quadrilateral ABCD, AB = DC and L B = L C.
Prove that AD I! BC.
99. If two medians of a A are equal, the A is isosceles.
NOTE.— Use Ex. 6, § 64.
1 00. If both pairs of opposite L. s of a quadrilateral are
equal, the quadrilateral is a ||gm.
101. Find the point on the base of a A such that the
difference of the _Ls from it to the sides is equal to a
given st. line.
102. Find the point on the base of a A such that the
sum of the _Ls from it to the sides is equal to a given st.
line.
103. Show that the three exterior /_s at A, C, E, in
the hexagon ABCDEF, are together less than the three
interior zs at B, D, F by two rt. /.s.
BOOK II
AREAS OF PARALLELOGRAMS AND TRIANGLES
70. A square unit of area is a square, each side of
which is equal to a unit of length.
Examples: — A square inch is a square each side of
which is one inch; a square centimetre is a square
each side of which is one centimetre.
The acre is an exceptional case.
71. A numerical measure of any area is the number
of times the area contains some unit of area.
A BCD is a rectangle one centimetre wide and five
centimetres long.
A D
B
This rectangle is a strip divided into five square
centimetres, and consequently the numerical measure
of its area in square centimetres is 5.
72. ABCD is a rectangle 3 cm. wide and 5 cm. long.
Ai 1 1 1 1 iD
This rectangle is divided into 5 strips of 3 sq. cm.
each, or into 3 strips of 5 sq. cm. each, and consequently
94 THEORETICAL GEOMETRY BOOK II
the measure of the area in square centimetres is 5 X 3
sq. cm., or 3 X 5 sq. cm.
Similarly, if the length of a rectangle is 2 '34 inches
and its breadth '56 of an inch, the orie-hundreth of
an inch may be taken as the unit and the rectangle
can be divided into 234 strips each containing 56
square one-hundreths of an inch. The measure of the
area then is 234 x 56 of these small squares, ten
thousand (100 X 100) of which make one square inch.
This method of expressing the area of a rectangle
may be carried to any degree of approximation, so
that in all cases the numerical measure of its area
is equal to the product of its length by its breadth.
In a rectangle any side may be called the base,
and then either of the adjacent sides is the altitude.
A rectangle, as A BCD, is commonly represented by
the symbol AB. BC, where AB and BC may be taken to
represent the number of units in the length and the
breadth respectively.
Or, if a be the measure of the base of a rectangle
and b the measure of its altitude, the area is ab.
In the case of a square, the base is equal to the
altitude, and if the measure of each be a, the area
is a2.
AREAS OF PARALLELOGRAMS AND TRIANGLES 95
THEOREM 1
The area of a parallelogram is equal to that of a
rectangle on the same base and of the same
altitude.
B . c
Hypothesis. — ABCD is a ||gm and EBCF a rectangle
on the same base BC and of the same altitude EB.
To prove that the area of the ||gm ABCD = the
area of rect. EBCF.
Proof. — v ED cuts the ||s AB, DC,
/. L EAB = L FDC. (1—9, p. 42.)
v ABCD is a || gm,
AB = CD. (1—20, p. 67.)
{L EAB = L FDC,
L AEB --= L DFC,
AB = DC,
.'. A AEB = A FDC, (1—14, p. 54.)
Figure EBCD - A EAB = ||gm ABCD,
Figure EBCD - A FDC = rect. EBCF;
and as equal parts have been taken from the same
area, the remainders are equal.
•*• ||gm ABCD = rect. EBCF.
Cor.— If a be the measure of the base of a ||gm
and 6 the measure of its altitude, the area, being
the same as that of a rect. of the same base and
altitude, = ab.
96 THEORETICAL GEOMETRY BOOK II
73.— Practical Exercises
1. Draw a ||gm having two adjacent sides 6-4 cm. and
7*3 cm. and the contained L 30°. Find its area.
2. Draw a ||gm having the two diagonals 4-8 cm. and
6-8 cm. and an L between the diagonals 75°. Find its
area.
3. The area of a ||gm is 50 sq. cm., one side is 10 cm.
and one L. is 60°. Construct the ||gm, and measure the
other side.
4. Draw a rectangle of base 7 cm. and height 4 cm.
On the same base construct a |jgm having the same area
as the rectangle and two of its sides each 65 mm. Measure
one of the smaller L s of the ||gm.
5. Make a ||gm having sides 10 and 7 cm. and one
L 60°. Make a rhombus equal in area to the ||gm and
having each side 10 cm. Measure the shorter diagonal of
the rhombus.
6. Make a rectangle 8 cm. by 5 cm. Construct a ||gm
equal in area to the rectangle and having two sides 7 cm.
and 8 cm. Construct a rhombus equal in area to the
|| gm and having each side 7 cm. Measure the shorter
diagonal of the rhombus.
7. Make a rhombus having each side 8 cm. and its area
50 sq. cm. Measure the shorter diagonal.
ANSWERS: — 1. 23-4 sq. cm. nearly. 2. 15-8 sq. cm. nearly.
3. 57 '7 mm. nearly. 4. 38° nearly. 5. 64 mm. nearly.
6. 64 mm. nearly. 7. 69 mm. nearly.
AREAS OF PARALLELOGRAMS AND TRIANGLES 97
THEOREM 2
Parallelograms on the same base and between
the same parallels are equal in area.
D K C F H E
Hypothesis. — A BCD, ABEF are ||gms on the same
base AB and between the same ||s AB, DE.
To prove that ||gm ABCD = ||gm ABEF.
Construction. — Draw AK, BH each J_ to both AB and
DE.
Proof. — v ilgm ABCD = rect. ABHK, (II — 1, p. 95.)
and ilgm ABEF = rect. ABHK,
.'. Ilgm ABCD = ligm ABEF.
THEORETICAL GEOMETRY
BOOK H
THEOREM 3
Parallelograms on equal bases and between the
same parallels are equal in area.
Hypothesis. — ABCD, EFGH are \\gma on the equal
bases AB, EF and between the same ||s AF, DG.
To prove that \\grn ABCD = j|gm EFGH.
Construction. — Draw AK, BL, EM, FN each _L to both
AF, DG.
Proof.— V AB = EF,
and AK = EM, (I — 20, p. 67.)
.'. rect. KB = rect MF.
But ||gm ABCD -rect. KB, (II— 1, p. 95.)
and !lgm EFGH = rect. MF,
/. Hgm ABCD = ||gm EFGH.
AREAS OF PARALLELOGRAMS AND TRIANGLES
99
74. Draw an acute- L d A ABC. Draw the _L from
A to BC. Draw through A, a st. line || BC. Show that
the JL distance between these || lines at any place =
the altitude of A ABC.
Draw an obtuse- L d A ABC, having the obtuse L.
at B. Draw the altitude AX.
Show that it falls without A
the A. Draw through A, a
st. line || BC. Show that
the distance between these
|| lines at any place = the X B C
altitude of the A.
Taking c as the vertex and AB as the base, draw
the altitude.
If a A is between two ||s, having its base in one
of the || s and its vertex in the other, its altitude is
the distance between the |js.
100
THEORETICAL GEOMETRY
BOOK II
THEOREM 4
The area of a triangle is half that of the rect-
angle on the same base and of the same altitude
as the triangle.
A D F E
Hypothesis. — ABC is a A and DBCE a rectangle on
the same base and of the same altitude BD.
To prove that area of A ABC = half that of rect.
DBCE.
Construction. — Through c draw CF || BA.
Proof. — V AC is a diagonal of ||grn ABCF,
.'. A ABC = half of ||gm ABCF. (I— 20, p. 67.)
But ||gm ABCF = rect. DBCE, (H.— 1, p. 95.)
/. A ABC = half of rect. DBCE.
Cor. — If a be the measure of the base of a A and
b the measure of its altitude, the measure of its area
is \ab.
75.— Practical Exercises
1. Draw a rt.-z_d A having the sides that contain the
right L 56 mm. and 72 mm. Find the area of the A.
2. Make a A ABC, having 6 = 6 cm., c = 8 cm., and
L A = 72°. Find its area.
3. Draw a A having yits sides 73 mm., 57 mm. and
48 mm. Find its area.
AREAS OF PARALLELOGRAMS AND TRIANGLES 101
4. Find the area of the A : a = 10 cm., L B = 42°,
L C = 58°.
5. The sides of a triangular field are 36 chains, 25 chains
and 29 chains. Draw a diagram and find the number of
acres in the field. (Scale: 1 mm. to the chain.)
6. Two sides of a triangular field are 41 and 38 chains
and the contained L is 70°. Find its area in acres.
ANSWERS: — 1. 20-16 sq. cm.; 2. 23 sq. cm. nearly;
3. 13-7 sq. cm. nearly; 4. 28'8 sq. cm.; 5. 36 ac. ; 6. 73
ac. nearly.
THEOREM 5
Triangles on the same base and between the
same parallels are equal in area.
X B Y <-
Hypothesis. — ABC, DBC are AS on the same base
BC and between the same ||s AD, BC.
To prove that A ABC = A DBC.
Construction. — Draw AX, DY j_ BC.
Proof.— A ABC = J rect. AX.BC. (II— ;4,p. 100.)
A DBC - \ rect. DY.BC.
But, V AX = DY,
/. rect. AX . BC - rect. DY. BC.
and .'. A ABC = A DBC.
102
THEORETICAL GEOMETRY
THEOREM 6
BOOK II
Triangles on equal bases and between the same
parallels are equal in area.
B« X' C E1 Y F
Hypothesis.— ABC, DEF are As en equal bases BC,
EF and between the same ||s AD, BF.
To prove that A ABC = A DEF.
Construction. — Draw AX, DY j_ BF.
Proof.— A ABC = J rect. AX.BC. (II.— 4, p. 100.)
A DEF = \ rect. DY.EF.
But, V BC = EF,
and AX = DY, (I.— 20, p. 67.)
/. rect. AX . BC = rect. DY . EF.
Hence, A ABC = A DEF.
Cor. i. — Triangles on equal bases and of the
same altitude are equal in area.
Cor. 2.— A median bisects the area of the triangle.
AREAS OF PARALLELOGRAMS AND TRIANGLES 103
THEOREM 7
If a parallelogram and a triangle are on the
same base and between the same parallels, the
parallelogram is double the triangle.
B C
Hypothesis. — ABCD is a ||gm and EBC a A on the
same base BC and between the same ||s AE, BC.
To prove that ||gm ABCD = twice A EBC.
Construction. — Draw BX, CY, EZ j_ BC and AE.
Proof. — 1| gm ABCD = rect. BX . BC. (II — 1, p. 95.)
A EBC = J rect EZ . BC. (II— 4, p. 100.)
But, v BX = EZ (1—20, p. 67.)
.*. rect. BX . BC = rect. EZ . BC.
And /. ||gm ABCD = twice A EBC.
76.— Exercises
VI. As ABC, DEF are between the same ||s AD and
BCEF, and BC > EF. Prove that A ABC > A DEF.
2. On the same base with a ||gm construct a rectangle
equal in area to the ||gm.
3. On the same base with a given ||gnv construct a ||gm
equal in area to the given ||gm, and having one of its
sides equal to a given st. line.
)( 4. Construct a rect. equal in area to a given ||gm, and
having one of its sides equal to a given st. line.
104 THEORETICAL GEOMETRY BOOK II
5. Make a ||gm with sides 5 cm. and 3 cm., and con-
tained L 125°. Construct an equivalent rect. having one
side 1 • 5 cm.
6. On the same base as a given A construct a rect.
equal in area to the A.
7. Construct a rect. equal in area to a given A, and
having one of its sides equal to a given st. line.
8. On the same base with a ||gm construct a rhombus
equal in area to the ||gm.
X' 9. Construct a rhombus equal in area to a given ||gm,
and having each of its sides equal to a given st. line.
10. On the same base with a given A, construct a
rt.- L d A equal in area to the given A.
p( 11. On the same base with a given A, construct an
isosceles A equal in area to the given A.
Vl2. If, in the ||gm ABCD, P be any point between AB,
CD produced indefinitely, the sum of the As PAB, PCD
equals half the ||gm ; and if P be any point not between
AB, CD, the difference of the As PAB, PCD equals half the
llgm-
Xl3. AB and ECD are two || st. lines; BF, DF are drawn
|| AD, AE respectively; prove that As ABC, DEF are equal
to each other.
14. On the same base with a given A, construct a A
equal in area to the given A, and having its vertex in a
given st. line.
/ 1 5. If two As have two sides of one respectively equal to
two sides of the other and the contained Ls supplementary,
the As are equal in area.
/ 16. ABCD is a ||gm, and P is a point in the diagonal
AC. Prove that A PAB = A PAD.
EXERCISES 105
\<>
17. P is a point within a ||gm A BCD. Prove that A
PAC equals the difference between As PAB, PAD.
X 18. In A ABC, BC and CA are produced to P and Q
/respectively, such that CP = one-half of BC, and AQ =
one-half of CA. Show that A QCP = three-fourths of A
ABC.
X 19. The medians BE, CD of the A ABC intersect at F.
^Show that A BFC = quadrilateral ADFE.
X 20. On the sides AB, BC of a A the || gms ABDE, CBFG
are described external to the A. ED and GF meet at H
and BH is joined. On AC the ||gm CAKL is described with
CL and AK || and = HB. Prove ||gm AL = ||gm AD -f ||gm
CF.
X 21. Two As are equal in area and between the same ||s.
Prove that they are on equal bases.
K 22. Of all As on a given base and between the same ||s,
the isosceles A has the least perimeter.
23. ABCD is a ||gm, and E is a point such that AE, CE
are respectively J_ and || to BD. Show that BE = CD.
X. 24. The side AB of ||gm ABCD is produced to E and DE
cuts BC at F. AF and CE are joined. Prove that
A AFE = A CBE.
25. In the quadrilateral ABCD, AB II CD. If AB = a,
CD = b and the distance between AB and CD = h, show
that the area of ABCD = \ h (a + b).
26. Two sides AB, AC of a A are given in length, find
the L A for which the area of the A will be greatest.
27. The medians AD, BE of A ABC intersect at G, and
CG is joined. Prove that the three lines AG, BG, CG trisect
the area of the A.
106 THEORETICAL GEOMETRY BOOK II
28. Bisect the area of a A by a st. line drawn through
a vertex.
29. Trisect the area of a A by two st. lines drawn
through a vertex.
X 30. Bisect the area of a A by a st. line drawn through
a given point in one of the sides.
31. Trisect the area of a A by two st. lines drawn
through a given point in one of the sides.
32. The area of any quadri-
lateral A BCD is equal to that
of a A having two sides and
their included L respectively
equal to the diagonals of the
quadrilateral and their in-
cluded L .
NOTE. — Draw PS and QR || BD, PQ and SR || AC. Join
SQ.
33. Prove that in a rhombus the distance between one
pair of opposite sides equals the distance between the other
pair.
34. 1 1 gins are described on the same base and between
the same ||s. Find the locus of the intersection of their
diagonals.
35. Prove that the area of a rhombus is half the pro-
duct of the lengths of its diagonals.
36. A BCD is a quadrilateral in which AB || CD, E is the
middle point of AD. Prove that A BEC - J quadrilateral
ABCD.
37. Divide a given A into seven equal parts.
AREAS OF PARALLELOGRAMS AND TRIANGLES 107
THEOREM 8
If two equal triangles are on the same side of a
common base, the straight line joining their vertices
is parallel to the common base.
A D
B
Hypothesis. — ABC, DBC are two equal As on the
same side of the common base BC.
To prove that AD || BC.
Construction. — Draw AX and DY _L BC.
Proof.— A ABC = J rect. BC . AX. (II — 4, p. 100.)
A DBC = J rect. BC.DY;
but A ABC = A DBC,
/. i rect. BC . AX - J rect. BC . DY
and hence AX = DY,
that is, AX and DY are both = and |) to each other
..ADHXY. (I— 19, p. 66.)
77. If, through any point E, in the diagonal AC of a
parallelogram BD, two straight lines PEG, HEK be
drawn parallel respectively to the sides DC, DA of the
parallelogram, D H C
A K B
the ||gms FK and HG are said to be parallelograms
108 THEORETICAL GEOMETRY BOOK II
about the diagonal AC, and the llgms DE, EB are
called the complements of the llgms FK, HG, which
are about the diagonal.
THEOREM 9
The complements of the parallelograms about
the diagonal of any parallelogram are equal to
each other.
K B
Hypothesis. — FK and HG are llgms about the diagonal
AC of the ||gm ABCD.
To prove that the complements DE, EB are equal to
each other.
Proof. — '.' AE is a diagonal of ||gm FK,
;. A AFE - A AKE. (1—20, p. 67.)
Similarly A HEC = A EGC.
.-. A AFE + A HEC = A AKE -f A EGC.
But, Y AC is a diagonal of ||gm ABCD
/. A ADC = A ABC.
/. A ADC - (A AFE + A HEC)
= A ABC - (A AKE + A EGC).
.'. llgm DE = ||gm EB.
EXERCISES 109
78. — Exercises
1. If two equal As be on equal segments of the same st.
line and on the same side of the line, the st. line joining
their vertices is II to the line containing their bases.
2. Through P, a point within the ||gm ABCD, EPF is
drawn || AB and GPH is drawn || AD. If ||gm AP = ||gm
PC, show that P is on the diagonal BD. (Converse of
Theorem 9.)
3. Two equal As ABC, DBC
are on opposite sides of the
same base. Prove that AD is
bisected by BC, or BC pro-
duced.
NOTE. — Produce DB making
BE = DB. Join EA, EC.
Give another proof of this proposition using J_s from
A and D to BC and II — 4, p. 100.
4. The median drawn to the base of a A bisects all st.
lines drawn il to the base and terminated by the sides, or
the sides produced.
5. P is a point within a A ABC and is such that A PAB
4- A PBC is constant. Prove that the locus of P is a st.
line || AC.
6. ||gms about the diagonal of a square are squares.
7. D, E, F are respectively the middle points of the sides
BC, CA, AB in the A ABC. Prove A BEF = A CEF and
hence that EF || BC.
8. In the diagram of II— 9, show that FK f| HG.
110 THEORETICAL GEOMETRY BOOK II
CONSTRUCTIONS
PROBLEM 1
To construct a parallelogram equal in area to a
given triangle and having one of its angles equal
to a given angle.
B H E C
Let ABC be the given A and D the given L.
It is required to construct a ||gm equal in area to
A ABC and having one L equal to L D.
Construction. — Through A draw AFG || BC. Bisect BC
at E. At E make L CEF = L D. Through C draw
CG || EF.
FC is the required ||gm.
Proof. — Draw any line HK J_ to the two [| st. lines.
HK is the common altitude of the ||gm FC and the
A ABC.
||gm FC = rect. EC. HK. (II— 1, p. 95.)
= \ rect. BC.HK, V EC = \ BC,
= A ABC. (II— 4, p. 100.)
CONSTRUCTIONS 111
PROBLEM 2
To construct a triangle equal in area to a given
quadrilateral.
B C E
Let A BCD be the given quadrilateral.
It is required to construct a A equal in area to
ABCD.
Construction. — Join AC. Through D draw DE || AC
and meeting BC produced at E. Join AE.
A ABE = quadrilateral ABCD.
Proof. — V D Ell AC,
/. A EAC = A DAC. (II — 5, p. 101.)
To each of these equals add A ABC.
Then A ABE = quadrilateral ABCD.
112 THEORETICAL GEOMETRY BOOK II
PROBLEM 3
To construct a triangle equal in area to a given
rectilineal figure.
A B G
Let the pentagon ABODE be the given rectilineal
figure.
Construction. — Join AD, BD. Through E, draw
EFII AD and meeting BA at F. Through C draw CG II BD
and meeting AB at G.
Join DF, DG.
A DFG = figure ABODE.
Proof.— v EF || AD,
. . A DFA = A DEA. ' (II— 5, p. 101.)
v CG II DB,
.. A DGB - A DOB.
. . A DFA + A DAB + A DBG
= A DEA + A DAB + A DOB ;
i.e., A DFG = figure ABODE.
By this method a A may be constructed equal in
area to a given rectilineal figure of any number of
sides; e.g., for a figure of seven sides, an equivalent
figure of five sides may be constructed, and then, as
in the construction just given, a A may be con-
structed equal to the figure of five sides.
CONSTRUCTIONS 113
PROBLEM 4
To describe a parallelogram equal to a given
rectilineal figure, and having an angle equal to
a given angle.
D L
A 0 B H
Let ABODE be the given rectilineal figure and F
the given L.
It is required to construct a ||gm = ABODE, and
having an Z_ = L F.
Construction. — Make A DMH equal in area to figure
ABODE. (II— Prob. 3, p. 112.)
Make ||gm LGHK = A DMH, and having L LGH =
L F. (II— Prob. l,p. 110.)
Then ||gm LGHK = figure ABODE, and has L LGH =
L F.
79.— Exercises
1. Construct a rect. equal in area to a given A.
2. Construct a rect. equal in area to a given quadri-
lateral.
3. Construct a quadrilateral equal in area to a given
hexagon.
4. On one side of a given A construct a rhombus equal
in area to the given A.
5. Construct a A equal in area to a given || gm, and
having one of its L. s = a given L. .
\
114 THEORETICAL GEOMETRY BOOK II
PROBLEM 5
To construct a triangle equal in area to a given
triangle and having one of its sides equal to a
given straight line.
B C E
Let ABC be the given A and D the given st. line.
It is required to make a A = A ABC and having
one side = D.
Construction. — From BC, produced if necessary, cut
off BE = D. Join AE. Through C draw CF II EA and
meeting BA, or BA produced at F. Join FE.
FBE is the required A.
Proof.— V FCIIAE,
/. A FCE = A AFC. (II— 5, p. 101.)
/. A FBC + A FCE = A FBC + A AFC,
i.e., A FBE = A ABC,
and side BE was made = D.
CONSTEUCTIONS 115
PROBLEM 6
On a straight line of given length to make a
parallelogram equal in area to a given triangle
and having an angle equal to a given angle.
B F C
Let ABC be the given A, E the given st. line and D
the given L.
It is required to make a ||gm equal in area to
A ABC, having one side equal in length to E, and one
L equal to D.
Construction. — From BC, produced if necessary, cut
off BF — E Join AF. Through C draw CG || FA meeting
BA, or BA produced, at G. Join GF. Bisect BG at H.
Through H draw HM || BC. At B make L CBL = L D.
Through F draw FM || BL.
LBFM is the required ||gm.
Proof.— Join HF.
As GAF, AFC are on the same base AF and have
the same altitude, /. they are equal. (II — 5, p. 101.)
To each of these equal As add the A ABF, and
A GBF = A ABC.
A GBF = twice A HBF, (II— 6, Cor. 2, p. 102.)
= ||gm LBFM, (II— 7, p. 103.)
.-. ||gm LBFM = A ABC.
Also L LBF = L D and side BF = E.
116 THEORETICAL GEOMETRY BOOK II
AREAS OF SQUARES
80. — A rectangle is said to be contained by two st.
lines when its length is equal to one of the st. lines,
and its breadth is equal to the other.
The symbol AB2 should be read: — "the square on
AB," and not "AB squared."
THEOREM 10
The square on the sum of two straight lines
equals the sum of the squares on the two straight
lines increased by twice the rectangle contained by
the straight lines.
A B C
Hypothesis. — AB, BC are the two st. lines placed in"
the same st. line so that AC is their sum.
To prove that
AC2 = AB2 -f BC2 + 2 . AB.BC.
Algebraic Proof
Proof. — Let a, b represent the number of units of
length in AB, BC respectively.
Area of the square on AC
= (a + &)2
= a2 + &2 + 2 ab
= area of square on AB + area of square on BC +
twice the area of the rectangle contained by AB,
and BC.
AREAS OF SQUARES
Geometric Proof
D L E
117
B b C
Construction. — On AC, AB, BC draw squares AC ED,
ABFG, BCKH. Produce BF to meet DE at L.
Proof.—
GD = AD - AG = AC - AB = BC, and GF = AB.
.-. GL = rect. AB.BC.
KE = CE - CK = AC - BC = AB, and HK = BC.
.-. HE = rect. AB.BC.
AC2 - AE
= AF + BK + GL + HE
= AB2 + BC2 + 2 AB.BC
118
THEORETICAL GEOMETRY
BOOK II
THEOREM 11
The square on the difference of two straight lines
equals the sum of the squares on the two straight
lines diminished by twice the rectangle contained
by the straight lines.
A~~ ~~C B
Hypothesis. — AB, BC are two st. lines, of which AB
is the greater, placed in the same st. line, and so that
AC is their difference.
To prove that
AC2 - AB2 + BC2 - 2 . AB . BC.
Algebraic Proof
Proof. — Let a, b represent the number of units of
length in AB, BC respectively.
Area of square on AC = (a — b)2
= & 4. 52 _ 2a&.
= the sum of the squares on AB and BC diminished by
twice the area of the rectangle contained by AB and
BC.
Geometric Proof
G a F
ab,-'
..a?-' E
*'"'
/
/
/(a-b)2
/ a-b
1
,ab/
/ b
i C
u
/ b2
Construction. — On AC, AB, BC draw the squares
AGED, ABFG, BCKH. Produce DE to meet BF at L.
AREAS OF SQUARES 119
Proof.— DG = AG - AD = AB - AC = BC, and DL = AB.
.-. DF = rect. AB . BC.
KE = KG + CE = BC -f AC = AB, and KH = BC.
/. KL = rect. AB . BC.
AC2 = AE
= AF-f KB - (DF+ KL)
= AB2 + BC2 - 2 AB. BC.
THEOREM 12
The difference of the squares on two straight
lines equals the rectangle of which the length is
the sum of the straight lines and the breadth is
the difference of the straight lines.
A, B are two st. lines, of which A > B.
To prove that the square on A diminished by the
square on B = the rect. contained by A -f B and A - B.
Proof. — Let a, b represent the number of units in
A and B respectively.
The difference of the squares on A and B
= a2 - 62
= (a + b) (a - b)
= the area of the rectangle -
contained by A 4- B and A - B.
120
THEORETICAL GEOMETRY
BOOK II
81.— Exercises
1. Draw a diagram illustrative of Theorem 12.
2. The square on the sum of three st. lines equals the
sum of the squares on the three st. lines increased by
twice the sum of the rectangles contained by each pair of
the st. lines.
Illustrate by diagram.
3. The sum of the squares on two unequal st. lines >
twice the rectangle contained by the two st. lines.
4. The sum of the squares on three % unequal st. lines >
the sum of the rectangles contained by each pair of the
st. lines.
5. Construct a rectangle equal to the difference of two
given squares.
6. If there be two st. lines AB and CD, and CD be
divided at E into any two parts, the rect. AB.CD = rect.
AB.CE + rect. AB.ED.
Let AB = p units of length
CE = q " " "
ED = r " " "
Area of AB . CD = p (q + r)
" AB.CE = pq
" " AB.ED = pr.
But p (q + r) = pq+pr.
:. AB . CD = AB . CE + AB . ED.
7. Give a diagram illustrating the identity (a + b)
(c + d) = ac + ad + be + bd, taking a, b, c, d to be
respectively the number of units in four st. lines.
8. C is the middle point of a „ i i
A R
st. line AB, and D is any other C D
point in the line. Prove :
B
EXERCISES 121
(1) AD.DB = AC2 - CD2;
(2) AD2 + DB2 = 2 AC2 -f 2 CD2.
(Let AC = CB = p, CD = q).
9. C is the middle point of
a st. line AB, and D is any
A ' O
point in AB produced. Prove : C B
(1) AD.DB = CD2 - AC2;
(2) AD2-f- DB2 = 2 AC2+ 2 CD'.
10. Draw diagrams to illustrate the four results in
exercises 8 and 9.
11. Draw a diagram illustrating the identity (a + b)2 —
(a - 6)2 = 4 ab.
12. If A, B, C, D be four points in order in a st. line,
AB . CD + AD . BC = AC . BD.
Illustrate by a diagram.
13. AB is a st. line in which C is any point. Prove
that AB2 = AB . AC + AB . CB.
14. Construct a A having two sides and the median
drawn to one of these sides equal to three given st. lines.
15. Construct a A having two sides and the median
drawn to the > third side equal to three given st. lines.
16. In a given ||gm inscribe a rhombus having one vertex at
a given point in a side of the ||gm.
122
THEORETICAL GEOMETRY
THEOREM 13
BOOK II
The square described on the hypotenuse of a
right-angled triangle is equal to the sum of the
squares on the other two sides.
Pl
/M;
f
/ i
/ 1
/
j
D
L
Hypothesis. — ABC is a A in which L ACB is a
rt. L, and AE, BG, CK are squares on AB, BC and CA.
To prove that AB2 = AC2 + BC2.
Construction. — Through c draw CL || AD.
Join KB. CD.
Proof. — V LS HCA, ACB, BCG are rt. LS,
.: LS HCB, ACG are st. LS.
and .*. HCB, ACG are st. lines.
L BAD = L KAC,
to each add L CAB,
then L CAD = L KAB.
T CA = KA
In As CAD, KAB, \ AD = AB
[ L CAD = L KAB
/. A CAD = A KAB (1—2, p. 16.)
EXERCISES 123
Y rect. ADLM and A CAD are on the same base
AD and between the same ||s CL, AD,
.'. rect. AL = twice A CAD. (II — 7, p. 103.)
Similarly, sq. HA = twice A KAB.
.*. rect. AL = sq. HA.
In the same manner, by joining CE and AF, it may
be shown that
rect. BL = sq. BG.
.'. rect. AL + rect. BL = sq. HA + sq. BG,
i.e., AB2 = AC2 -f BC2.
82. Many proofs have been given for this important
theorem. Pythagoras (570 to 500 B.C.) is said by
tradition to have been the first to prove it, and from
that it is commonly called the Theorem of Pythagoras,
or the Pythagorean Theorem. The proof given above is
attributed to Euclid (about 300 B.C.). An alternative
proof is given in Book IV.
83. — Exercises
1. Draw two st. lines 5 cm. and 6 cm. in length.
Describe squares on both, and make a square equal in
area to the two squares. Measure the side of this last
square and check your result by calculation.
2. Draw three squares having sides 1 in., 2 in. and
2^ in. Make one square equal to the sum of the three.
Check by calculation.
3. Draw two squares having sides 1J in. and 2J in.
Make a third square equal to the difference of the first
two. Check by calculation.
4. Draw two squares having sides 9 cm. and 6 cm.
Make a third square equal to the difference of the first
two. Check your result by calculation.
124
THEORETICAL GEOMETRY
BOOK II
5. Draw any square and one of its diagonals. Draw
a square on the diagonal and show that it is double
the first square.
6. Draw a square having each side 4 cm. Draw a second
square double the first. Measure a side, and check by
calculation.
7. Draw a square having one side 45 min. Draw a
second square three times the first. Measure its side,
and check by calculation.
8. Draw three lines in the ratio II 2 1 3. Draw squares
on the lines, and divide the two larger so as to show that
the squares are in the ratio 1 ; 4 ; 9.
9. Draw a st. line j/ 2 in. in length.
10. Draw a st. line -j/Sin. in length.
1 1. Draw a st. line y 5 in. in length.
12. Draw any rt.-Z-d A. Describe equilateral As on
the three sides. Find the areas of the As and compare
that on the hypotenuse with the sum of those on the
other two sides.
13.
AB is one inch in length, L B a rt. z_ , BC is one inch
BD is cut off = AC, BE = AD, BF = AE, BG = AF, etc.
Show that BD = -j/2 in-> BE - -j/3 in., BF = y/4 == 2
in., BG =
in. etc.
EXERCISES • 125
14. Construct a square equal to half a given square.
15. If a ± be drawn from the vertex of a A to the base,
the difference of the squares on the segments of the base =
the difference of the squares on the other two sides.
Hence, prove that the altitudes of a A pass through one
point.
16. A is a given st. line. Find another st. line B, such
that the difference of the square on A and B may be equal
to the difference of two given squares.
y**\7. If the diagonals of a quadrilateral cut at rt. L s, the
sum of the squares on one pair of opposite sides equals the
sum of the squares on the other pair.
18. The sum of the squares on the diagonals of a rhom-
bus equals the sum of the squares on the four sides.
y£^19. Five times the square on the hypotenuse of a rt.-z.d
A equals four times the sum of the squares on the medians
drawn to the other two sides.
20. In an isosceles rt.-z_d A the sides have the ratios
1 : 1 : >/ 2.
21. If the angles of a A be 90°, 30', 60°, the sides have
the ratios 2 : 1 : |/ 3.
W 22. Divide a st. line into two parts such that the sum
of the squares on the parts equals the square on another
given st. line. When is this impossible?
C*23. In the st. line AB produced find a point C such that
the sum of the squares on AC, BC equals the square on a
given st. line.
i^-24. Divide a given st. line into two parts such that the
square on one part is double the square on the other part.
)^25. ABCD is a rect., and P is any point. Show that
PA2 -f- PC2 = PB2 + PD2.
26. ABC is a A rt.- L d at A. E is a point on AC and
F is a point on AB. Show that BE2 + OF2 = EF2 + BC2.
126 • THEORETICAL GEOMETRY BOOK II
27. If two rt.- L d As have the hypotenuse and a side of
one respectively equal to the hypotenuse and a side of the
other, the As are congruent.
28. The square on the side opposite an acute L of a A
is less than the sum of the squares on the other two sides.
29. The square on the side opposite an obtuse L of a A
is greater than the sum of the squares on the other two
sides.
30. Construct a square that contains 20 square inches.
31. In the diagram of II — 13, show that KB, CD cut
at rt. L. s.
32. In the diagram of II — 13, if KD be joined, show
that A KAD = A ABC.
33. In the diagram of II — 13, the distance of E from
AC = AC + CB.
34. ABC is an isosceles rt.- L d A in which C is the
rt. L. CB is produced to D making BD = CB. J_s to AB,
BD at A, D respectively meet at E. Prove that AE = 2
AB.
AREAS OF SQUARES 127
THEOREM 14
(Converse of Theorem 13)
If the square on one side of a triangle is equal
to the sum of the squares on the other two sides,
the angle contained by these sides is a right angle,
B C E F
Hypothesis. — ABC is a A in which BC2 = AB2 + AC2.
To prove that L A is a rt. /_.
Construction. — Make a rt. L D and cut off DE = AB,
DF = AC
Join EF.
BC2 = AB2 + AC2 (Hyp.)
= DE2 -f DF2
= EF2 (V D is a rt. /_). (11—13, p. 122.)
.. BC = EF.
IAB - DE,
AC- DF,
BC = EF,
.'. L A = L D. (1—4, p. 22.)
.'. L A is a rt. L.
128 THEORETICAL GEOMETRY BOOK II
84.— Exercises
1. The sides of a A are 3 in., 4 in. and 5 in. Prove
that it is a rt.- L. d A.
2. The sides of a A are 13 mm., 84 mm. and 85 mm.
Prove that it is a rt.- L d A.
3. In the quadrilateral ABCD, AB2 -f CD2 = BC2 + AD2.
Prove that the diagonals AC, BD cut at rt. L. s.
4. If the sq. on one side of a A be less than the sum
of. the squares on the other two sides, the L contained by
these sides is an acute L . (Converse of § 83, Ex. 28.)
5. State and prove a converse of § 83, Ex. 29.
6. Using a tape-measure, or a knotted cord, and Ex. 1,
draw a st. line at rt. L s to a given st. line.
7. Show that, if the sides of a A are represented by
m? + n*, m2 - nz, 2 mn, where ra and n are any numbers,
the A is rt.-_d.
Use this result to find numbers representing the sides
of a rt.- L. d A.
85. Definition. — If a perpendicular be drawn from a
given point to a given straight line, the foot of the
perpendicular is said to be the projection of the
point on the line.
From the point A the j_ AX is drawn to the line
BC.
P
B X C
The point X is the projection of the point A on the
st. line BC.
EXERCISES 129
86. Definition.— If from the ends of a given
straight line perpendiculars be drawn to another
given straight line, the segment intercepted on the
second straight line is called the projection of the
first straight line on the second straight line.
AB is a st. line of fixed length and CD another st.
line. AE, BF are drawn J_ CD.
CE F D C !/ FDCE FD
EF is the projection of AB on CD.
87.— Exercises
1. Show that a st. line of fixed length is never less than
its projection on another st. line. In what case are they
equal1? In what case is the projection of one st. line on
another st. line just a point?
2 ABC is a A having a = 36 mm., b = 40 mm. and
c = 45 mm. Draw the A and measure the projection of
AB on BC. (Ans. 23'9 mm. nearly.)
3. ABC is a A having a = -j cm., 6 = 7 cm., c =. 10
cm. Draw the A and measure the projection of AB on
BC. (Ans. 76 mm.)
130
THEORETICAL GEOMETRY
THEOREM 15
BOOK II
In an obtuse-angled triangle, the square on the
side opposite the obtuse angle equals the sum of
the squares on the sides that contain the obtuse
angle increased by twice the rectangle contained
by either of these sides and the projection on that
side of the other.
Hypothesis. — ABC is a A in which L C is obtuse,
and CD is the projection of CA on CB.
To prove that AB2 = AC2 + BC2 + 2 BC. CD.
Proof.— -: ADB is a rt. Z,
. . AB2 = BD2 + AD2. (H—13, p. 122.)
V BD = BC + CD,
'. BD2 = BC' + CD2+ 2 BC.CD. (II— 10, p. 116.)
.. AB2 = BC2 + CD2 + 2 BC. CD -f AD2,
v ADC is a rt. L ,
.. CD2+ AD2 = AC2.
.'. AB2 = AC2 + BC2+ 2 BC.CD.
But
EXERCISES 131
THEOREM 16
In any triangle, the square on the side opposite
an acute angle is equal to the sum of the squares
on the sides which contain the acute angle
diminished by twice the rectangle contained by
either of these sides and the projection on that
side of the other.
B D C D B C
Hypothesis. — ABC is a A in which Z C is acute, and
CD is the projection of CA on CB.
To prove that AB2 = AC2 + BC2 - 2 BC.CD.
Proof. — v ADB is a rt. Z,
/. AB2 = BD2 + AD2. (H—13, p. 122.)
V BD is the difference between BC and CD,
;. BD2 = CD2 + BC2 - 2 BC.CD. (II— 11, p. 118.)
.'. AB2 = CD2 + BC2 - 2 BC.CD -f AD2.
But, Y ADC is a rt. z_,
.'. CD2 4- AD2 = AC2.
.'. AB2 = AC2 + BC2 - 2 BC.CD.
88.— Exercises
J 1 ABC is a A having C an z of 60°. Show that sq.
on AB = sq. on BC + sq. on AC - rect. BC.AC.
2. ABC is a A having C an z of 120°. Show that sq.
on AB = sq. on BC 4- sq. on AC + rect. BC.AC.
'
132 THEORETICAL GEOMETRY BOOK II
1 3. ABC is a A, CD the projection of CA on CB, and CE
the projection of CB on CA. Show that rect. BC.CD =
rect. AC.CE.
1 4. In any A the sum of the squares on two sides equals
twice the square on half the base together with twice the
square on the median drawn to the base.
NOTE. — Draw a J_ from the vertex to the base, and use II—
15 and 11—16.
\ 5. In any quadrilateral the sum of the squares on the
four sides exceeds the sum of the squares on the diagonals
by four times the square on the st. line joining the middle
points of the diagonals.
What does this proposition become when the quadri-
lateral is a Hgm1?
V 6. ABC is a A having a = 47 mm., b = 62 mm., and
c = 84 mm. D, E, F are the middle points of BC, CA,
AB respectively. Calculate the lengths of AD, BE and CF.
Test your results by drawing and measurement.
7. The squares on the diagonals of a quadrilateral are
together double the sum of the squares on the st. lines
joining the middle points of opposite sides.
8. If the medians of a A intersect at G,
AB2 -f BC2 -f CA2 = 3 (GA2 + GB2 -f GC2).
t 9. C is the middle point of a st. line AB. P is any
point on the circumference of a circle of which C is the
centre. Show that PA2 + PB2 is constant.
10. Two circles have the same centre. Prove that the
sum of the squares of the distances from any point on the
circumference of either circle to the ends of the diameter
of the other is constant.
I 11. The square on the base of an isosceles A is equal to
twice the rect. contained by either of the equal sides and
the projection on it of the base.
EXERCISES
133
12. Prove II — 13 from the following construction: —
Draw two squares, A BCD,
AEFG, having AD, AE in
the same st. line.
Cut off GH and EK each
= AB.
Join FH, HC, CK, KF.
13. If two sides of a A
be unequal, the median
drawn to the shorter side
is greater than the median
drawn to the longer side.
14. If, from any point P within A ABC, _Ls PX, PY, PZ
be drawn to BC, CA, AB respectively,
BX2 4- CY2 4- AZ2 = CX2 4- AY2 4- BZ2.
15. D, E, F are the middle points of BC, CA, A B respec-
tively in A ABC. Prove that
3 (AB2 4- BC2 4- CA2) = 4 (AD2 4- BE2 4~ CF2).
16. G is the centroid of A ABC, and P is any point.
Show that
PA2 4- PB2 4- PC2 = AG2 4- BG2 4- CG2 + 3 PG2.
17. Find the point P in the plane of the A ABC such
that the sum of the squares on PA, PB, PC may be the
least possible.
18. Check the results in Exs. 2 and 3, §87, by calculation.
19. If, in II — 15, the obtuse L. becomes greater and
greater and finally becomes a st. L. , what does the theorem
become ?
20. If, in the diagram of II — 16, the L. C becomes more
and more acute and finally the point A comes down to the
line BC, what does the theorem become ?
134 THEORETICAL GEOMETRY BOOK II
Miscellaneous Exercises
1. If a quadrilateral be bisected by each of its diagonals,
it is a || gm.
2. If any point P in the diagonal AC of the ||gm ABCD
be joined to B and D, the ||gm is divided into two pairs
of equal As.
3. The diagonals of a ||gm divide the ||gm into four
equal parts.
4. If two sides of a quadrilateral are || to each other,
the st. line joining their middle points bisects the area of
the quadrilateral.
5. If two sides of a quadrilateral are || to each other, the
st. line joining their middle points passes through the inter-
section of the diagonals.
6. If P is any point in the side AB of ||gm ABCD, and
PC, PD are joined,
A PAD + A PBC = A PDC.
7. Prove that the following method of bisecting a quadri-
lateral by a st. line drawn through one of its vertices is
correct: — Let ABCD be the quadrilateral. Join AC, BD.
Bisect BD at E. Through E draw EF || AC and meeting
BC, or CD, at F. Join AF. AF bisects the quadrilateral.
NOTE.— Join AE, and EC.
8. If the diagonals of ||gm ABCD cut at O, and P is
any point within the A AOB, A CPD = A APB -f A APC +
A BPD.
NOTE.— Join PC.
9. ABC is an isosceles A having AB = AC, and D is a
point in the base BC, or BC produced. Prove that the
difference between the squares on AD and AC = rect.
BD.DC.
MISCELLANEOUS EXERCISES 135
10. P, Q, R, S are respectively the middle points of the
sides AB, BC, CD, DA in the quadrilateral ABCD. Prove
that AB2 + CD2 + 2 PR2 = CB2 + DA2 -f 2 QS2.
11. BY _L AC and CZ J_ AB in A ABC. Prove that
BC2 = rect. AB.BZ -f rect. AC.CY.
12. L, M, N are three given points, and PQ a given st.
line. Construct a rhombus ABCD, having its angular points
A, C lying on the line PQ, and its three sides AB, BC, CD
(produced if necessary) passing through L, M, N respectively.
13. Through D the middle point of the side BC of A
ABC a st. line XDY is drawn cutting AB at X and AC
produced through C at Y. Prove A AXY > A ABC.
14. From the vertex A of A ABC draw a st. line
terminated in BC and equal to the average of AB and AC.
15. AB and CD are two equal st. lines that are not in
the same st. line. Find a point P such that A PAB = A
PCD.
Show that, in general, two such points may be found.
16. EF drawn || to the diagonal AC of ||gm ABCD meets
AD, DC, or those sides produced, in E, F respectively.
Prove that A ABE = A BCF.
17. Construct a rect. equal to a given square and such
that one side equals a given st. line.
18. Find a point in one of two given intersecting st.
lines such that the perpendiculars drawn from it to both
the given lines may cut off from the other a segment of
given length.
19. In the diagram of II — 9, if BD, BE and DE be
drawn, ||gm FK - ||gm HG = 2 A EBD.
20. ABC is an isosceles A in which C is a rt. L , and
the bisector of L A meets BC at D. Prove that CD =
AB - AC.
136 THEORETICAL GEOMETRY < BOOK II
21. Place a st. line of given length between two given
st. lines so as to be || a given st. line.
22. Describe a A = a given |jgm and such that its base
= a given st. line, and one L at the base = a given L .
23. Construct a ||gm equal and equiangular to a given
||gm, and such that one side is equal to a given st. line.
24. Construct a ||gm equal and equiangular to a given
||gm, and such that its altitude is equal to a given st. line.
25. ABCD is a quadrilateral. On BC as base construct
a || gm equal in area to ABCD, and having one side along
BA.
26. Squares ABDE, ACFG have a common z_ A, and
A, B, C are in the same st. line. AH is drawn J_ BG
and produced to cut CE at K. Prove that EK = KC.
27. Make a rhombus ABCD in which L A = 100°. A circle
described with centre A and radius AB cuts BC, CD at E,
F respectively. Prove that AEF is an equilateral A.
28. A st. line AB is bisected at C and divided into two
unequal parts at D. Prove that AD2 + DB2 = 2AD.DB +
4 CD2.
29. ABCD is a quadrilateral in which AB || CD. Prove
that
AC2 + BD2 = AD2 + BC2 + 2 AB.CD.
30. Trisect a given llgm by st. lines drawn through one
of its angular points.
31. The base BC of the A ABC is trisected at D, E.
Prove that
AB2 + AC2 = AD2 + AE2 + 4 DE2.
32. ACB, ADB are two rt.- L d As on the same side of
the same hypotenuse AB, and AX, BY are j_ CD produced.
Prove that
XC2 4- CY2 = XD2 4- DY3.
MISCELLANEOUS EXERCISES 137
33. ABC is an isosceles A, and XY is || BC and termi-
nated in AB, AC. Prove
BY2 = CY2 + BC.XY.
34. Any rect. = half the rect. contained by the diagonals
of the squares on two of its adjacent sides.
35. ABCD is a ||gm in which BD — AB. Prove that
BD2 -f 2 BC2 = AC2.
36. A rect. B DEC is described on the side BC of a A ABC.
Prove that
AB2 + AE2 = AC2 + AD2.
37. BE, CD are squares described externally on the sides
AB, AC of a A ABC. Prove that
BC2 + ED2 = 2 (AB2 + AC2).
NOTE. — Draw EX, CY J_ DA, AB respectively, and rotate
A ABC to the position in which AB coincides with AE.
38. ABC is a A in which AX JL BC, and D is the middle
point of BC. Prove that the difference of the squares on
AB, AC = 2BC.DX.
39. BC is the greatest and AB the least side in A ABC.
D, E, F are the middle points of BC, CA, AB respectively;
and X, Y, Z are the feet of the _Ls from A, B, C to the
opposite sides. Prove that CA.EY = AB.FZ + BC.DX.
40. ABCD is a rect. in which E is any point in BC and
F is any point in CD. Prove that ABCD = 2 A AEF +
BE.DF.
41. A and B are two fixed points. Find the position of
a point P such that PA2 + PB2 may be the least possible.
42. From a given point A draw three st. lines AB, AC, AD
respectively equal to three given st. lines, and such that
B, C, D are in the same st. line and BC = CD.
43. Find the locus of a point such that the sum of the
squares on its distances from two given points is constant.
138 THEORETICAL GEOMETRY BOOK II
• <
44. Find the locus of a point such that the difference of
the squares on its distances from two given points is
constant.
45. A BCD is a ||gm, P any point in BC, and Q any
point in A P. Prove that A BQC = A PQD.
46. A BCD is a quadrilateral having AB || CD, and AB +
CD = BC. Prove that the bisectors of z_s B and C inter-
sect on AD.
47. ABC is a A in which /_ A is a rt. /_ , and AB >
AC. Squares BCDE, CAHF, ABGK are described out-
wardly to the A. Prove that
DG2 - EF2 = 3 (AB2 - AC2).
48. In the hypotenuse AB of a rt.-^d A ACB, points D
and E are taken such that AD = AC and BE = BC. Prove
that
DE2 = 2 BD. AE.
49. A st. line is 8 cm. in length. Divide it into two
parts such that the difference of the squares on the parts
= 5 sq. cm.
50. A and B are two given points and CD is a given st.
line. Find a point P in CD such that the difference of the
squares on PA and PB may be equal to a given rectangle.
51. AD is a median of the acute- ^d A ABC; DX _L
AB, DY JL AC. Prove that
BA . AX + CA.AY = 2 AD2.
52. Find a point P within a given quadrilateral KLMN
such that A PLM = A PMN ^ A PNK.
53. ABC is an isosceles A in which AB = AC. AP || BC.
Prove that the difference between PB2 and PC2 equals
2 AP.BC.
54.' If the sum of the squares on the diagonals of a
quadrilateral be equal to the sum of the squares on the
sides, the quadrilateral is a ||gm.
MISCELLANEOUS EXERCISES 139
55. D is a point in the side BC of a A ABC such that
AB2 + AC2 = 2 AD2 + 2 BD2. AX JL BC. Prove that either
BD = DC, or 2 DX = BC.
56. A BCD is a ||gm, and P is a point such that PA2 -f-
PC2 = PB*+ PD2. Prove that ABDC is a rectangle.
57. A, B, C, D are four fixed points. Find the locus of
a point P such that PA2 + PB2 + PC2 + PD2 is constant.
58. A, B, C, D are four fixed points. Find the locus of
a point P such that PA2 -f PB2 = PC2 + PD2.
59. D and E are taken in the base BC of A ABC so that
BD = EC. Through D, E st. lines are drawn || AB and AC
forming two ||gms with AD, AE as diagonals. Prove the
|| gins equal in area.
60. A st. line EF drawn || to the diagonal AC of a ||gm
A BCD meets AB in E and BC in F. Prove that BD bisects
the quadrilateral DEBF.
61. ABC is an isosceles rt.-/.d A in which AB = AC. E
is taken in AB and D in AC produced such that EB = CD.
Prove that A EAD < A ABC.
62. L and M are respectively the middle points of the
diagonals BD and AC of a quadrilateral A BCD. ML
is produced to meet AD at E. Prove that A EBC =
half the quadrilateral.
63. DE is || BC the base of A ABC, and meets AB, AC
at D, E respectively. DE is produced to F making DF =
BC. Prove that A AEF - A BDE.
64. Construct the minimum A which has a fixed
vertical /_ , and its base passing through a fixed point
situated between the arms of the L .
65. BE, BD are the bisectors of the interior and exterior
^s at B in the A ABC.AE _L BE and CD _L BD. AE
and CD intersect at F. Prove that rect. BEFD = A ABC.
140 THEORETICAL GEOMETRY , BOOK II
66. ABCD is a square. St. lines drawn through A and
D make with BC produced in both directions the A EFG.
EX J. FG. Prove that BC(EX + FG) = 2 A EFG.
67. The A ABC is rt.- L d at C, and the bisectors of L s
A and B meet at E. ED J_ AB. Prove that rect. AD.DB
'= A ABC.
68. Calculate the area of an equilateral A of which the
side is 2 inches.
69. If the side of an equilateral A is a inches, show
that its area is a ^ sq. in.
4
70. Calculate the side of an equilateral A of which the
area is 10 sq. cm.
71. Construct a A having two sides 4 cm. and 4-5 cm.,
and the area 7 sq. cm.
Show that there are two solutions.
72. M is a point in the side QR of A PQR such that
QM = 2 MR. Prove that PQ2 + 2 PR2 = 3 PM2 + 6 MR2.
73. The rectangle contained by the two segments of a
st. line is a maximum when the st. line is bisected. (Use
Ex. 8 (1), §81.)
74. The sum of the squares on the two segments of a
st. line is a minimum when the st. line is bisected. (Use
Ex. 8 (2), §81.)
BOOK III
THE CIRCLE
89. A definition of a circle was given in § 32, and
from the explanation given in § 66 we may take the
following alternative definition of it : —
A circle is the locus of the points that lie at a
fixed distance from a fixed point.
90. As the centre of a circle is a point equally
distant from the two ends of any chord of the circle,
the three following statements follow at once from
1—22, p. 78 :—
(a) The straight line drawn from the centre of a
circle perpendicular to a chord bisects the chord.
(b) The straight line drawn from the centre of a
circle to the middle point of a chord is perpen-
dicular to the chord.
(c) The right bisector of a chord of a circle passes
through the centre of the circle.
As an exercise the pupil should give independent
proofs of theorems (a), (b) and (c).
141
142 THEORETICAL GEOMETRY < BOOK III
THEOREM 1
If from a point within a circle more than two
equal straight lines are drawn to the circumference,
that point is the centre.
Hypothesis. — P is a point within the circle ABC such
that PA = PB = PC.
To prove that P is the centre of the circle.
Construction. — Join AB, BC, and from P draw
PD _L AB and PE _L BC.
Proof.— ': PA = PB,
.*. P is in the right bisector of AB. (I — 22, p. 78.)
And .'. PD produced is the locus of the centres of
all circles through A and B.
/. the centre of the circle ABC is somewhere in PD.
In the same manner it may he shown that the centre
of the circle ABC is somewhere in PE.
But P is the only point common to PD arid PE.
.% P is the centre of circle ABC.
CONSTRUCTIONS 143
CONSTRUCTIONS
PROBLEM 1
To find the centre of a given circle.
Let DEF be the given circle.
Construction. — From any point D on the circum-
ference draw two chords DE, DF.
Draw the right bisectors of DE, DF meeting at O.
O is the centre of circle DEF.
Join OF, OE, OO OP
Proof. — Y O is on the right bisector of DE,
.'. OE = OD. (1—22, p. 78.)
Similarly OD — OF.
Y OE = OD = OF,
/. O is the centre of the circle. (Ill— 1, p. 142.)
91. Definitions. — If a circle passes
through all the vertices of a rectilineal
figure, it is said to be circumscribed
about the figure.
Four points so situated that a circle
may be described to pass through all of them are said
to be concyclic.
144 THEORETICAL GEOMETRY BOOK III
If the four vertices of a quadrilateral are on the
circumference of the same circle, it is said to be a
cyclic quadrilateral.
The centre of a circle circumscribed about a
triangle is called the circumcentre of the triangle.
PROBLEM 2
To circumscribe a circle about a given triangle.
Let PQR be the given A.
Construction. — Draw the right bisectors of PQ, PR
meeting at O.
V O is on the right bisector of PQ.
/. OP = OQ. (I— 22, p. 78.)
Similarly OP = OR.
.'. OP = OQ = OR,
And a circle described with centre O and radius OP
will pass through Q and R, and be circumscribed about
the A.
EXERCISES 145
92.— Exercises
1. Through a given point within a circle draw a chord
that is bisected at the given point.
2. Complete a circle of which an arc only is given.
3. Circumscribe a circle about a given square.
4. Circumscribe a circle about a given rectangle.
5. Describe a circle with a given centre to cut a given
circle at the ends of a diameter.
6. The locus of the middle points of a system of || chords
in a circle is a diameter of the circle.
7. If two circles cut each other, the st. line joining their
centres bisects their common chord at rt. z_s.
8. If each of two equal st. lines has one extremity on
one of two concentric circles, and the other extremity on
the other circle, the st. lines subtend equal L. s at the
common centres.
9. A st. line cuts the outer of two concentric circles at
E, F; and the inner at G, H. Prove that EG = FH.
10. A st. line cannot cut a circle at more than two points.
11. Two chords of a circle cannot bisect each other
unless both are diameters.
12. A circle cannot be circumscribed about a ||gm unless
the ||gm is a rectangle.
13. A st. line which joins the middle points of two
|| chords in a circle is J_ to the chords.
14. If two circles cut each other, a st. line through a
point of intersection, || to the line of centres and termi-
nated in the circumferences, is double the line joining the
centres.
146 THEORETICAL GEOMETRY < BOOK III
15. If two circles cut each other, any two || st. lines
through the points of intersection, and terminated by the
circumferences, are equal to each other.
16. If two circles cut each other, any two. st. lines
through one of the points of intersection, making equal
L s with the line of centres and terminated by the
circumferences, are equal to each other.
THEOREM 2
Chords that are equally distant from the centre
of a circle are equal to each other.
B c
Hypothesis. — ABC is a circle of which P is the centre
and AB, CD are two chords such that the J_s PE, PF
from P to A B, CD respectively are equal to each other.
To prove that AB = CD.
Construction. — Join AP, CP.
Proof. — Rotate A PFC about point P making PF
fall on PE.
V PF = PE,
.'. point F falls on point E.
V L PFC = L PEA,
.*. FC falls alonof EA.
CONSTRUCTIONS 147
hence, v PC is a radius and
/. C remains on the circumference,
C must fall on A.
.'. FC coincides with EA,
and .'. FC = EA,
But CD = 2CF,
and AB = 2 AE,
.'. CD = AB.
THEOREM 3
In a circle any chord which does not pass
through the centre is less than a diameter.
Hypothesis. — In the circle FGH, GH is a chord which
does not pass through the centre and FK is a diameter.
E is the centre.
To prove that GH < FK.
Construction. — Join EG, EH.
Proof.— V GE = EF and EH = EK,
. . GE + EH = FK.
V GEH is a A,
.'. GH < GE -f EH. (1—16, p. 59.)
And .*. GH < FK.
148
THEORETICAL GEOMETRY
BOOK III
THEOREM 4
Of two chords in a circle the one which is nearer
to the centre is greater than the one which is more
remote from the centre.
Hypothesis. — P is the centre of a circle ABC, and
AB. CD are two chords such that PE, the distance of
AB from the centre, is less than PF, the distance of
CD from the centre.
To prove that AB > CD.
Construction. — Join PA, PC.
Proof.— V PEA is a rt. L,
.', AE2 + EP2 = AP2. (11—13, p. 122.)
Similarly CF2 + FP2 = CP2.
But V AP = CP,
AP2= CP2.
And . . AE2 + EP2 = CF2 -f FP2.
EP < PF,
EP2<PF2.
And .*. AE2>CF2,
AE > CF.
But AB = 2 AE,
and CD = 2 CF,
AB > CD.
CONSTRUCTIONS
149
THEOREM 5
(Converse of Theorem 4)
If two chords of a circle are unequal, the greater is
nearer to the centre than the less.
(11—13, p. 122.)
Hypothesis. — Chord GH > chord KL, and PE, PF
are respectively perpendiculars from the centre P to
GH, KL.
To prove that PE < PF.
Construction. — Join PG, PK.
PEG is a rt. /_ ,
GE2 + EP2 = GP2.
PF2 + FK2 = PK-.
GE2+ EP2 = PF2+ FK2.
GH = 2 GE and KL = 2 KF,
GH > KL,
GE > KF.
GE?> KF2.
EP2 < PF2.
EP < PF.
Proof.—
Similarly
But
And also
Hence,
And
150 THEORETICAL GEOMETRY BOOK III
(
93.— Exercises
1. If two chords of a circle are equal to each other,
they are equally distant from the centre. (Converse of
Theorem 2.)
2. A chord 6 cm. in length is placed in a circle of
radius 4 cm. Calculate the distance of the chord from the
centre.
3. A chord a inches long is placed in a circle of radius
b inches. Find an algebraic expression for the distance of
the chord from the centre.
4. In a circle of radius 5 cm. a chord is placed at a
distance of 3 cm. from the centre. Calculate the length
of the chord.
5. Through a given point within a circle draw the
shortest chord.
6. In a circle of radius 4 cm., a point P is taken at
the distance 3 cm. from the centre. Calculate the length
of the shortest chord through P.
7. The length of a chord 2 cm. from the centre of a
circle is 5 '5 cm. Find the length of a chord 3 cm. from
the centre. Verify your result by measurement.
8. In a circle of radius 5 cm., two || chords of lengths
8 cm. and 6 cm. are placed. Find the distance between
the chords. Show that there are two solutions.
9. ACB is a diameter, and C the centre of a circle. D
is any point on AB, or on AB produced, and P is any
point on the circumference except A and B. Show that DP
is intermediate in magnitude between DA and DB.
10. O is the centre of a circle, and P is any point. If
two st. lines be drawn through P, cutting the circle, and
EXERCISES
151
making equal L s with PO, the chords intercepted on these
lines by the circumference are equal to each other.
11. O is the centre of a circle, and P is any point. On
two lines drawn through P chords AB, CD are intercepted
by the circumference. If the L made by AB with PO >
L made by CD with PO, the chord AB < chord CD.
12. From any point in a circle which is not the centre
equal st. lines can be drawn to the circumference only in
pairs.
13. Find the locus of the middle points of chords of a
fixed length in a circle.
14 K and L are two fixed points. Find a point P on
a given circle such that KP2 + LP2 may be the least
possible.
15. Chords equally distant from the centre of a circle
subtend equal L s at the centre.
16. The nearer to the centre of two chords of a circle
subtends the greater L at the centre.
ANSWERS : — 2, 26*5 mm. nearly ; 4, 8 cm.; 6, 5-3 c.m. nearly;
7, 32 mm. nearly ; 8, 1 cm. or 7 cm.
152 THEORETICAL GEOMETRY , BOOK III
ANGLES IN A CIRCLE
THEOREM 6
The angle which an arc of a circle subtends at
the centre is double the angle which it subtends at
any point on the remaining part of the circumference.
Hypothesis. — ABC is an arc of a circle, D the centre,
and E any point on the remaining part of the
circumference.
To prove that L ADC = 2 Z A EC.
Construction. — Join ED and produce ED to any
point F.
Proof.—
In both figures : —
In A DAE, Y DA = DE
.*. L DAE = L DEA (1—3, p. 20).
'.* ADF is an exterior L of A ADE,
.'. L ADF = Z DAE + L DEA (I— 10, p. 45.)
= 2 Z DEA.
Similarly Z CDF = 2 L DEC.
ANGLES IN A CIRCLE 153
In Fig. 1 :—
Z ADF = 2 Z DEA
Z CDF = 2 Z DEC,
adding, Z ADC = 2(Z DEA + Z DEC)
= 2 Z AEC.
In Fig. 2 :—
Z CDF = 2 Z DEC,
Z ADF = 2 Z DEA,
subtracting, Z ADC = 2(Z DEC - Z DEA).
= 2 Z AEC.
94. Definitions. — The figure bounded by an arc of
a circle and the chord which joins the ends of the arc
is called a segment of a circle.
A CD F
ABC, DEF are segments of circles.
A semi-circle is a particular case of a segment.
An arc is called a major arc or a minor arc
according as it is greater or less than half the
circumference.
A segment is called a major segment or a minor
segment according as the arc of the segment is a
major or a minor arc.
154 THEORETICAL GEOMETRY ? BOOK III
95. Definitions. — If the ends of a chord of a seg-
ment are joined to any point on the arc of the segment,
the angle between the joining lines is called an angle
in the segment.
ABC is an Z in the segment ABC, and DEF is an
Z in the segment DEF. DGF is also an Z in the
segment DEF.
96. Definitions. — An angle which is greater than
two right angles but less than four right angles is
called a reflex angle.
A straight line starting from the position OX and
rotating in the direction opposite to that of the hands
of a clock to the position OY, in either diagram,
traces out the reflex angle XOY.
ANGLES IN A CIRCLE
155
The figure bounded by two radii of a circle, and
either of the arcs intercepted by the radii is called a
sector of the circle.
ABC, DEFG are sectors of circles.
BAG is the L of the sector ABC, and the reflex L
EDG is the Z of the sector DEFG.
156
THEORETICAL GEOMETRY
THEOREM 7
BOOK III
Angles in the same segment of a circle are equal
to each other.
Hypothesis. — ABC, ADC are two z_s in the same
segment ABDC.
To prove that L ABC = L ADC.
Construction. — Find E the centre of the circle.
Join AE, EC.
Proof. — The L A EC at the centre and the Ls ABC
and ADC at the circumference are subtended by the
same arc,
.'. L ABC = i L AEC, (HI— 6, p. 152.)
and Z ADC = i Z AEC,
.*. Z ABC = Z ADC.
Alternative statement of the preceding theorem : —
The angle in a given segment is constant in
magnitude for all positions of the vertex of the
angle on the arc of the segment.
ANGLES IN A CIRCLE
157
THEOREM 8
(Converse of Theorem 7 )
The locus of all points on one side of a straight
line at which the straight line subtends equal
angles is the arc of a segment of which the straight
line is the chord.
a
Hypothesis. — AB is a st. line, and C one of the
points. Circumscribe a circle about the A ACB.
To prove that arc ACB is the locus of all points on
the same side of AB at which AB subtends Zs equal
to Z ACB. V
Construction. — Take any other point D on arc ACB,
E any point within the segment, and F any point
without the segment.
Join AD, DB, AE, EB, AF, FB.
Proof.— Then z ADB = L ACB. (HI— 7, p. 156.)
Produce AE to meet arc ACB at G. Join BG.
Y AEB is an exterior Z of A EGB,
.*. Z AEB > Z AGB; (I— 10, Cor., p. 45.)
but Z AGB = Z ACB, (III— 7, p. 156.)
/. L AEB > Z ACB ;
In a similar manner it may be shown that
Z AFB < Z ACB;
and consequently arc ACB is the locus.
158 THEORETICAL GEOMETRY ( BOOK III
97. Definition : — If the three angles of one triangle
are respectively equal to the three angles of another
triangle, the triangles are said to be similar. ) \ \
98. There are two conditions implied when figures
are said to be similar: not only are the angles of one
respectively equal to the angles of the other, but a
certain relationship must exist between the lengths of
the sides of the two figures. For triangles, it will be
shown in Book IV that, if one of these conditions is
given, the other is also true. For figures of more
than three sides this is not the case, and a definition
including both conditions must be given. (See § 131.)
The symbol ||| may be used for the word similar, or
for " is similar to."
99.— Exercises
*f 1. Prove Theorem 6 when the arc is half the circumference.
2. Construct a circular arc on a chord of 3 inches and
having the apex 3 inches from the chord. Calculate the
radius of the circle.
3. If the chord of an arc is a inches, and the distance
of its apex from the chord b inches, show that the radius
of the circle is "' + ^
8 o
4. Two chords AOB, COD, intersect at a point O within
the circle. Show that AOC, BOD are similar As. BOC,
AOD are also similar As. Read the segments that contain
the equal L s.
5. ABC is a A inscribed in a circle, and the bisector of
L A meets the circumference again at D. Show that the
st. line drawn from D J_ BC is a diameter.
EXERCISES 159
f^ 6. A circle is divided into two segments by a chord
equal to the radius. Show that the L in the major
segment is 30° and that in the minor segment is 150°.
7. The locus of the vertices of the rt. L s of all rt.- L d
As on the same hypotenuse is a circle.
8. Prove Theorem 6 when the arc is greater than half
the circumference.
/ 9. PQR is a A inscribed in a circle. The bisector of
L P cuts QR at D and meets the circle at E. Prove that
A PQD HI A PER.
/ 10. DPQ and EPQ are two fixed circles, and D, P and E
are in the same st. line. The bisector of L DQE meets
DE at F. Show that the locus of F is an arc of a circle.
*""11. If the diagonals of a quadrilateral inscribed in a
circle cut at rt. L s, the _L from their intersection on any
side bisects the opposite side.
1 2. If the diagonals of a quadrilateral inscribed in a circle
cut at rt. L. s, the distance of the centre of the circle from any
side is half the opposite side.
13. If the diagonals of a quadrilateral inscribed in a
circle cut each other at rt. L s, the L s which a pair of
opposite sides of the quadrilateral subtend at the centre of
the circle are supplementary.
*-- 14. XYZ, XYV are two equal circles, the centre of each
being on the circumference of the other. ZXV is a st.
line. Prove that YZV is an equilateral A.
15. EFGH is a quadrilateral inscribed in a circle and
EF = GH. Prove that EG = FH.
Vl6. ABCD is a quadrilateral inscribed in a circle; the
diagonals AC, BD cut at E ; F the centre of the circle is
within the quadrilateral. Prove that Z AFB+ L CFD =
2 Z AEB.
160 THEORETICAL GEOMETRY BOOK III
THEOREM 9
The angle in a semi-circle is a right angle.
Hypothesis. — ABC is an L in the semi-circle ABC,
of which D is the centre.
To prove that ABC is a rt. L .
Proof. — The Z ABC at the circumference, and the
st. L ADC at the centre, would each subtend the
same arc, if the circle were complete.
.'. Z ABC = i Z ADC. (HI— 6, p. 152.)
= a rt. L.
ANGLES IN A CIRCLE
THEOREM 10
161
(a) The angle in a major segment of a circle is
acute.
(b) The angle in a minor segment of a circle is
obtuse.
Fig 1 Fig.2
(a) Hypothesis. — ACB is an Z in a major segment of
a circle. (Fig. 1.)
To prove that Z ACB is acute.
Construction. — Join A and B to the centre D.
Proof. — Z ACB at the circumference and Z ADB at
the centre stand on the same arc,
/. Z ACB = i Z ADB. (Ill— 6, p. 152.)
But L ADB is < a st. Z.
.*. Z ACB is acute.
(b) Hypothesis. — ACB is an Z in a minor segment of
a circle. (Fig. 2.)
To prove that Z ACB is obtuse.
Construction. — Join A and B to the centre D.
Proof.—
Z ACB = i the reflex Z ADB. (Ill— 6, p. 152.)
• Z ACB is obtuse.
162 THEORETICAL GEOMETRY BOOK III
100.— Exercises
1. A circle described on the hypotenuse of a rt.-z_d A as
diameter passes through the vertex of the rt. L . (Converse
of III-9).
2. Circles described on two sides of a A as diameters,
intersect on the third side, or the third side produced.
Where is the point of intersection when the circles are
described on the equal sides of an isosceles A 1
3. LM is a st. line and L a point from which it is
required to draw a JL to LM.
Construction. — With a con-
•v _ M venient point P as centre de-
scribe a circle to pass through
L and cut LM at D. Join DP,
and produce DP to cut the
circle at E. Join LE.
Prove LE JL LM.
4. EF, EG are diameters of two circles FEH, GEH re-
spectively. Show that FHG is a st. line.
5. ST is a diameter of the circle SVT. A circle is
described with centre S and radius ST. Show that any
chord of this latter circle drawn from T is bisected by the
circle SVT.
6. Chords of a given circle are drawn through a given
point. Find the locus of the middle points of the chords
when the given point is (a) on the circumference, (b) within
the circle, (c) without the circle.
7. F is any point on the arc of a semi-circle of which
DE is a diameter. The bisectors of L s FED, FDE meet
at P. Find the locus of P.
ANGLES IN A CIRCLE
163
8. F is a point on the arc of a semi-circle of which DE
is a diameter. FG _L DE. Show that the As FDG, FEG,
FDE are similar.
9. PQRS is a st. line and circles described on PR, QS
as diameters cut at E. Prove that L PEQ = L RES.
THEOREM 11
If a quadrilateral is inscribed in a circle, its
opposite angles are supplementary.
D
Hypothesis. — A BCD is a quadrilateral inscribed in a
circle.
To prove that z_A+Z_C=2rt. Zs.
Construction.— Find the centre E. Join BE, ED.
Proof. — Z BED at the centre and L C at the cir-
cumference are subtended by the same arc BAD.
/. Z C = i L BED. (HI— 6, p. 152.)
Similarly Z A -- \ reflex L BED.
Hence L A + L C = \ the sum of the two Z s
BED at the centre = J of 4 rt. Z.S
= 2 rt. Zs.
164 THEORETICAL GEOMETRY BOOK III
THEOREM 12
If the opposite angles of a quadrilateral are sup-
plementary, its vertices are concyciic.
C E
Hypothesis. — ABCD is a quadrilateral in which
L A+ L C = 2 rt. La.
To prove that A, B, C, D are on the circumference
of a circle.
Construction. — Draw a circle through the three
points A, B, D. On this circumference and on the
side of BD remote from A take a point E. Join
BE, ED.
Proof. — *.' ABED is a quadrilateral inscribed in a
circle,
.-. L A+ L E = 2 rt. zs; (III— 11, p. 163.)
but L A -f- L C = 2 rt. Ls. (Hyp.)
:. ZA-f^E-Z_A + LC,
and /. L E = L C.
Consequently, as C, E are on the same side of BD,
the circle BADE passes through C. (Ill — 8, p. 157.)
EXERCISES 165
101.— Exercises
1. If one side of an inscribed quadrilateral be produced,
the exterior L thus formed at one vertex
equals the interior L at the opposite
vertex of the quadrilateral.
State and prove the converse.
2. From a point O without a circle
two st. lines OAB, OCD are drawn
cutting the circumference at A, B, C, D.
Show that As OBC, OAD are similar, and that As OAC,
OBD are similar.
3. If a || gm be inscribed in a circle, the ||gm is a rect.
4. A, D, C, E, B are five successive points on the cir-
cumference of a circle; and A, B are fixed. Show that
the sum of the L s ADC, CEB is the same for all positions
of D, C, E.
5. A circle is circumscribed about an equilateral A.
Show that the L in each segment outside the A is an L
of 120°.
6. A scalene A is inscribed in a circle. Show that the
sum of the L s in the three segments outside the A is
360°.
7. A quadrilateral is inscribed in a circle. Show that
the sum of the L. s in the four segments outside the quadri-
lateral is 540°.
8. P is a point on the diagonal KM of the ||gm KLMN.
Circles are described about PKN and PLM. Show that LN
passes through the other point of intersection of the circles.
9. A circle drawn through the middle points of the sides -^^
of a A passes through the feet of the JLs from the vertices
to the opposite sides.
fc/fO. If the opposite sides of a quadrilateral inscribed in a
circle be produced to meet at L and M, and about the AS
166 THEORETICAL GEOMETRY BOOK III
so formed outside the quadrilateral circles be described in-
tersecting again at N, then L, M, N are in the same st.
line.
t 11. In a A DEF, DX JL EF and EY J_ DF. Prove that
L XYF = L DEF.
, 12. PQRS, PQTV are circles and SPV, RQT are st. lines.
ProVe^that SR || VT.
i/ 13. The st. lines that bisect any L. of a quadrilateral
inscribed in a circle and the opposite exterior L meet on
the circumference.
14. XYZ is a A ; YD JL ZX, and DE J_ XY; ZF _L XY
and FG _]_ ZX. Show that EG || YZ.
15. EGO, FGD are two circles with centres H, K re-
spectively. EGF is a st. line. EH, FK meet at P. Show
that H, K, D, P are coney clic.
16. KL, MN are two || chords in a circle; KE, NF two
_L chords in the same circle. Show that LF J_ ME.
1 7. The bisectors of the L s formed by producing the
opposite sides of a quadrilateral inscribed in a circle are
J_ to each other.
18. HKM, LKM are two circles, and HKL is a st. line.
HM, LM cut the circles again at E, F respectively, and
HF cuts LE at G. Show that a circle may be circum-
scribed about MEGF.
19. PQRS is a quadrilateral and the bisectors of the ^.s
Pi Q ; Q» R » RJ S ; S, P meet at four points. Show that a
circle may be circumscribed about the quadrilateral thus
formed.
20. EF is the diameter of a semi-circle and G, H any
two points on its arc. EH, FG cut at K and EG, FH cut
at L. Show that KL J_ EF.
ANGLES IN A CIRCLE 167
21. DE is the diameter, O the centre and P any point
on the arc of a semi-circle. PM j_ DE, Show that the
bisector of L MPO passes through a fixed point.
22. PQR is a A and PDQ, PFQ are two circles cutting
PR at D, F and QR at E, G. Prove that DE || FG.
THEOREM 13
If two angles at the centre of a circle are equal
to each other, they are subtended by equal arcs.
H
A
Hypothesis. — AKC, DKF are equal z_s at the centre
K of the circle ACD.
To prove that arc A EC equals arc DGF.
Construction. — Draw the diameter HKL bisecting
L CKD.
Proof. — Suppose the circle to be folded along the
diameter HKL, and the semi-circle HFL will coincide
throughout with the semi-circle HAL.
V L LKD = L LKC,
.*. KD falls along KC;
and /. D falls on C.
V L DKF = L CKA,
/. KF falls along KA;
and /. F falls on A.
/. the arc DGF coincides with the arc CEA.
• arc DGF = arc CEA.
168 THEORETICAL GEOMETRY - BOOK III
102. - Exercises
1. If two arcs of a circle be equal to each other, they
subtend equal La at the centre. (Prove either by indirect
demonstration, or by the construction and method used in
III— 13.)
2. If two La at the circumference of a circle be equal to
each other, they are subtended by equal arcs.
3. If two arcs of a circle be equal to each other, they
subtend equal As at the circumference.
4. In equal circles equal L s at the centres (or circum-
ferences) stand on equal arcs.
5. In equal circles equal arcs subtend equal L s at the
centres (or circumferences).
6. If two arcs of a circle (or of equal circles) be equal,
they are cut off by equal chords.
7. If two chords of a circle be equal to each other, the
major and minor arcs cut off by one are respectively equal
to the major and minor arcs cut off by the other.
8. If two sectors of a circle have equal La at the
centre, the sectors are congruent.
9. Bisect a given arc of a circle.
10. Parallel chords of a circle intercept equal arcs.
Show also that the converse is true.
I 11. If two equal circles cut one another, any st. line
drawn through one of the points of intersection will meet
the circles again at two points which are equally distant
from the other point of intersection.
1 2. The bisectors of the opposite L s of a quadrilateral
inscribed in a circle meet the circumference at the ends of
a diameter.
1 3. If two L s at the centre of a circle be supplementary,
the sum of the arcs on which they stand is equal to half
the circumference.
1 4. If any number of L s be in a segment, their bisec
tors all pass through one point.
TANGENTS AND CHORDS 169
TANGENTS AND CHORDS
103. Definitions. — Any straight line which cuts a
circle is called a secant.
A straight line which, however far it may be pro-
duced, has one point on the circumference of a circle,
and all other points without the circle is called a
tangent to the circle.
A tangent is said to touch the circle.
The common point of a tangent and circle, that is,
the point where the tangent touches the circle, is
called the point of contact.
A'BC is a secant drawn to the circle BCF from the
point A.
D^E is a tangent to the circle BCF, touching the
circle at the point of contact F.
If the secant ABC rotate about the point A until
the two points B, C where it cuts the circle coincide
at G, the secant becomes a tangent having G for the
point of contact.
170
THEORETICAL GEOMETRY
THEOREM 14
BOOK III
The radius drawn to the point of contact of a
tangent is perpendicular to the tangent.
B A
Hypothesis. — ABF is a tangent to the circle CBD at
the point B, O is the centre and OB the radius drawn
to the point of contact.
To prove that OB is ± AF.
Construction. — From any point A, except B, in AF
draw a secant AE cutting the circle in C and D. Join
OC, OD.
Proof.— V OD = OC,
.'. L ODC = L OCD. (I — 3, p. 20.)
But, St. L EDO = st. L DC A,
.-. L ODE = L OCA.
Rotate AE about A until it coincides with AF. As
AE rotates about A the Zs ODE, OCA are continually
equal to each other and finally L ODE becomes L OBF
and L OCA becomes L OBA.
.'. L OBF - L OBA.
and .*. OB J_ AF.
EXERCISES 171
Cor. I.— Only one tangent can be drawn at any
point on the circumference of a circle.
Y only one st. line can be JL to the radius at that
point.
Hence, also: — The straight line drawn perpen-
dicular to a radius at the point where it meets
the circumference is a tangent.
Cor. 2.— The perpendicular to a tangent at its
point of contact passes through the centre of the
circle.
.*. only one st. line can be _L to the tangent at that
point.
Cor. 3.— The perpendicular from the centre on a
tangent passes through the point of contact.
Y only one _l_ can be drawn from a given external
point to a given st. line.
104.— Exercises
1. Draw a tangent to a given circle from a given point
on the circumference.
2. Describe a circle with its centre on a given st. line DE
to pass through a given point P in DE and touch another
given st. line DF.
3. Find the locus of the centres of all circles that touch
a given st. line at a given point.
4. Describe a circle to pass through a given pojnt and
touch a given st. line at a given point.
5. Tangents at the ends of a 'diameter are ||.
.* /
172 THEORETICAL GEOMETRY , BOOK III
6. C is any point on the tangent of which A is the
point of contact. The st. line from C to the centre O cuts
the circumference at B. AD is JL OC. Show that BA
bisects the L DAC.
7. Find the locus of the centres of all circles which touch
two given || st. lines.
j/'S. Draw a circle to touch two given || st. lines and pass
through a given point between the ||s. Show that two
such circles may be drawn.
J 9. To a given circle draw two tangents, each of which
is || to a given st. line.
10. To a given circle draw two tangents, each of which
is J. to a given st. line.
11. Give an alternative proof for III — 14 by supposing
the radius OB drawn to the point of contact of the tan-
gent ABF not J_ to AF and drawing OG J. AF.
12. Two tangents to a circle meet each other. Prove that
they are equal to each other.
13. EF is a diameter of a circle and EG is a chord.
EH is a chord bisecting the L FEG. Prove that the
tangent at H is ± EG.
1 4. Draw a circle to touch a given st. line at a given
point and have its centre on another given st. line.
15. Draw a tangent to a given circle making a given L.
with a given st. line.
Show that, in general, four such tangents may be drawn.
CONSTRUCTION 173
CONSTRUCTION
PROBLEM 3
To draw a tangent to a given circle from a given
point without the circle.
N
C
Let ABC be the given circle, and P the given point.
It is required to draw a tangent from P to the
circle ABC.
Join p to the centre O. Bisect OP at D. With
centre D and radius DO, describe a circle cutting the
circle ABC at A and C. Join PA, PC.
Either PA or PC is a tangent to the given circle.
Join OA.
OAP is an L in a semi - circle, and is .*. a
rt. L. (Ill— 9, p. 160.)
.'. PA is a tangent. (Ill— 14, Cor. 1, p. 171.)
In the same manner it may be shown that PC is a
tangent.
THEORETICAL GEOMETRY BOOK III
105. Definition.— The straight line joining the
points of contact of two tangents to a circle is called
the chord of contact of the tangents.
B
BC is the chord of contact of the tangent AB, AC.
106. — Exercises
1. Draw a circle of radius 4 cm. Take a point 9 cm.
from the centre of the circle. From this point draw two
tangents to the circle. Measure the length of each tangent
and check your result by calculation.
2. Draw a circle of radius 5 cm. Mark a point 7 cm.
from the centre. From this point draw two tangents to
the circle and measure the L between the tangents. (91°
nearly.)
3. Draw a circle with a radius of 3 cm. Mark any
point A on the circumference, and from this point draw a
tangent AB 4 cm. long. Measure the distance of B from
the centre and check your result.
4. Draw a circle with 43 mm. radius. Draw ^any st.
line through the centre, and find a point, in this line,
from which the tangent to the circle will be 5 cm. in
EXERCISES 175
length. Measure the distance of the point from the centre
and check your result.
5. Mark two points A and B 7 cm. apart. Draw^ two
st. lines from A such that the length of the perpendicular
from B to either of them is 4 cm.
6. Draw a circle of radius 6 cm. Mark a point P 4
cm. from the centre. Draw a chord through P such that
the perpendicular from the centre to the chord is 3 cm. in
length. Measure the length of the chord and check your
result by calculation.
7. Draw a circle of radius 36 mm. Mark any point P
without the circle. Draw a st. line from P such that the
chord cut off on it by the circle is 4 cm. in length.
8. Draw a circle of radius 47 mm. Mark a point P 4
cm. from the centre. Draw two chords through P, each
of which is 65 mm. in length.
9. If from a point without a circle two tangents be
drawn, the st. line drawn from this point to the centre
bisects the chord of contact and cuts it at rt. La.
\/fO. If a quadrilateral be circumscribed about a circle,
the sum of one pair of opposite sides equals the sum of
the other pair.
Xll. Through a given point draw a st. line, such that
the chord intercepted on the line by a given circle is
equal to a given st. line.
/1 2. If a || gm be circumscribed about a circle, the llgm
is a rhombus.
13. If two tangents to a circle be ||, their chord of
contact is a diameter.
17(5 THEORETICAL GEOMETRY BOOK III
^ 14. If two || tangents to a circle be cut by a third
tangent to the circle at A, B ; show that AB subtends a
rt. L at the centre.
t
i"- 15. If a quadrilateral be circumscribed about a circle,
the L. s subtended at the centre by a pair of opposite sides
are supplementary.
16. To a given circle draw two tangents containing an
L equal to a given L .
17. Find the locus of the points from which tangents
drawn to a given circle are equal to a given st. line.
^18. Find a point P in a given st. line, such that the
tangent from P to a given circle is of given length. What
is the condition that this is possible ?
1^ 19. E is a point outside a circle the centre of which is
D. In DE produced find a point F, such that the length
of the tangent from F may be twice that of the tangent
from E.
20. Two tangents, LM, LN are drawn to a circle; P is
any point on the circumference outside the A LMN. Prove
that L LMP + L LNP is constant.
21. Find the L between the tangents to a circle from a
point whose distance from the centre is equal to a
diameter.
22. Show that all equal chords of a given circle touch
a fixed concentric circle.
23. From a given point without a circle draw a st. line
such that the part intercepted by the circle subtends a
rt. L at the centre.
TANGENTS AND CHORDS
177
THEOREM 15
If at one end of a chord of a circle a tangent is
drawn, each angle between the chord and the tangent
is equal to the angle in the segment on the other side
of the chord.
E A D
Hypothesis. — AB is a chord and BAD a tangent to the
circle ABC.
To prove that Z DAB = Z ACB and that Z EAB =
Z AHB.
Construction. — From A draw the diameter AOC. Join
BC. Join any point H in the arc AHB to A and B.
Proof. — v ABC is an Z in a semi-circle,
:. ABC is a rt. L. . (Ill — 9, p. li
/. Z BAG + ^ BCA = a rt. L (I— 10, p. 45.)
= z CAD. (Ill— 14, p. 170.)
Take away the common Z BAG,
.. Z BAD = Z ACB,
= Z in the segment ACB.
v AHBC is an inscribed quadrilateral,
.-. Z H + Z C = a st. L (HI— 11, p. 163.)
= si.L DAE
But Z C = Z BAD.
V Z BAE = Z H
= Z in the segment AHB.
178 THEORETICAL GEOMETRY BOOK III
THEOREM 15
(Alternative Proof)
If at one end of a chord of a circle a tangent
is drawn, each angle between the chord and the
tangent is equal to the angle in the segment on
the other side of the chord.
c
H
Hypothesis. — AB is a chord and EAD a tangent to
the circle ABC.
To prove that L DAB = L ACB, and that L EAB =
L AHB.
Construction. — In arc AFC take any point F. Join
CF, and draw the line FAG.
Proof.— '.' AFCB is an inscribed quadrilateral,
/. L FCB is supplementary to L FAB,
(III— 11, p. 163.)
But, L BAG is supplementary to L FAB.
/. L BAG = L FCB.
These Zs are equal however near F is to A.
Let F move along the circumference towards A and
finally coincide with A.
EXERCISES 179
The line FAG rotates about the point A and finally
coincides with EAD. The L GAB becomes L DAB and
L FOB becomes L ACB.
.'. L DAB = L ACB.
V L EAB is supplementary to L DAB,
and, L AHB is supplementary to L ACB.
(Ill— 11, p. 163.)
/. L EAB = L AHB.
107.— Exercises
1. AB is a chord of a circle and AC is a diameter. AD
is JL to the tangent at B. Show that AB bisects the
L DAC.
2. Two circles intersect at A and B. Any point P on
the circumference of one circle is joined to A and B and
the joining lines are produced to meet the circumference of
the other circle at C, D. Show that CD is || to the tangent ;
at P.
3. LMN is a A. Show how to draw the tangent at L
to the circumscribed circle, without finding the centre of
this circle.
4. If either of the La which a st. line, drawn through one
end of a chord of a circle, makes with, the chord is equal to
the L in the segment on the other side of the chord, the
st. line is a tangent. (Converse of III — 15.)
5. The tangent at a point P on a circle meets the chord
MN produced through N, at Q. Prove L Q = L PNM -
L PMN.
6. A tangent drawn || to a chord of a circle bisects the
arc cut off by the chord.
180 THEORETICAL GEOMETRY BOOK III
FGE, HKE are two circles, and FEH, GEK two st.
lines. Prove that FG, KH meet at an /_ which = the L.
between the tangents to the circles at E.
8. G is the middle point of an arc EGF of a circle.
Show that G is equidistant from the chord EF and the
tangent at E.
9. A st. line EF is trisected in G, H, and an equilateral
A PGH is described on GH. Show that the circle FGP
touches EP.
10. D, E, F are respectively the points of contact of the
sides MN, NL, LM of a A circumscribed about a circle.
DG, EH are respectively J_ EF, DF. Prove GH || LM.
11. The tangent at L to the circumscribed circle of A
LMN meets MN produced at D, and the internal and
external bisectors of the L MLN meet MN at E, F
respectively. Prove that D is the middle point of EF.
12. GEF, HEF are two circles and GEH is a st. line.
The tangents at G, H meet at K. Show that K, G, F, H
are concyclic.
13. Points P, Q are taken on two st. lines LM, LN so
that LP + LQ = a given st. line. Prove that the circle
PLQ passes through a second fixed point.
• 14. E, F, G, H are the points of contact of the sides
XY, YZ, ZV, VX of a quadrilateral circumscribed about a
circle. If X, Y, Z, V are concyclic, show that EG J_ FH.
15. XYZV is a quadrilateral inscribed in a circle, and
XZ, YV cut at E. Prove that the tangent at E to the
circle XEY is || ZV.
16. F is the point of contact of a tangent EF to the
circle FGH. GK drawn || EF meets FH, or FH produced,
EXERCISES 181
at K. Show that the circle through G, K, H touches FG
at G.
17. If from an external point P a tangent PT and a
secant PMN be drawn to a circle, the As PTM, PNT are
similar.
18. Use III — 15 to prove that the tangents drawn to
circle from an external point are equal.
a
19. From an external point T a tangent TR and a
secant TQP through the centre are drawn to a circle. Prove
that L T + 2 L TRQ = a rt. L.
20. The tangents OT, OS from a fixed point O to a
given circle contain an L of x degrees. A third tangent
is drawn to the circle at any point on the minor arc TS.
Show that the portion of this tangent intercepted by OT
and OS subtends an L of (90 - §) degrees at the centre.
Show that if the moving point be taken on the major
arc TS, the L at the centre will be (90 + |) degrees.
182 THEORETICAL GEOMETRY BOOK III
CONSTRUCTIONS
PROBLEM 4
On a given straight line to construct a segment,
containing an angle equal to a given angle.
D
F\
Let AB be the given st. line, and C the given z.
Construction. — Make L BAF = L C.
Draw AE J_ AF.
Draw the right bisector of AB and produce it to
cut AE at E.
V E is in the right bisector of AB, it is equidistant
from A and B. (1—22, p. 78.)
With centre E and radius EA describe the arc ADB.
ADB is the required arc.
Proof. — V AF is j_ AE,
.". AF is a tangent to the circle ADB.
(Ill— 14, Cor. 1, p. 171.)
V AB is a chord drawn from the point of contact
of the tangent AF,
/. Z in segment ADB - L FAB. (Ill — 15, p. 177.)
But, L FAB = L C, (Const.)
:. Z in segment ADB = L C.
EXERCISES 133
PROBLEM 5
From a given circle to cut off a segment con-
taining an angle equal to a given angle.
L E.
Let LMN be the given circle, and D the given Z.
Construction. — Draw a tangent LE to the given
circle.
At L make the L ELN = L D.
LMN is the required segment.
Proof. — v LE is a tangent, and LN a chord,
/. Z in segment LMN = L NLE.
(Ill— 15, p. 177.)
But, L NLE = L D. (Const.)
:. Z in segment LMN = L D.
108.— Exercises
1. On st. lines each 4 cm. in length, describe segments
containing /. s of (a) 45°, (6) 150°, (c) 72°, (d) 116°. (Use
the protractor for (c) and (d).)
2. On a given base construct an isosceles A with a
given vertical L. .
184 THEORETICAL GEOMETRY BOOK III
3. Divide a circle into two segments such that the L. in
one segment is (a) twice, (b) three times, (c) five times, (d)
seven times the L in the other segment.
4. Construct two As ABC], ABC2 on the same base
AB - 4 cm., having L ACiB = L AC2B = 50°, and ACj =
AC2 = 5 cm.
Prove that L ABC^ + L ABC2 = 2 rt. L s.
5. Construct a A LMN having LM = 5 cm., L N = 110°,
and the median from N = 2 cm.
Measure the greatest and least values the median from
N could have, with LM = 5 cm., and L N = 110°.
6. Construct a A having its base 5 cm., its vertical L
70°, and its altitude 3 cm.
7. Construct a A XYZ, having XY = 4 cm., L Z = 40°,
and XZ + ZY = 10 cm.
8. Construct a A XYZ, having XY = 6 cm., L Z = 50°
and XZ - ZY = 4 cm.
9. Through a given point draw a st. line to cut off
from a given circle a segment containing an L equal to a
given L .
CONSTRUCTIONS 185
PROBLEM 6
In a given circle to inscribe a triangle similar to
a given triangle.
L
F
Let LMN be the given circle, and DEF the given A.
Construction. — Draw a radius OL of the circle.
Make L LON = 2 L E, and Z LOM = Z 2 F.
Join LM, MN, NL.
LMN is the required A.
, Join OM, ON.
Proof. — v L LON at the centre and L LMN at the
circumference stand on the same arc.
.'. L LON = 2 L LMN, (III— 6, p. 152.)
But L LON = 2 L E, (Const.)
:. L LMN = L E.
Similarly L LNM = L F.
V L LMN = L E,
and A LNM = L F,
.'. Z. MLN = z. D. (I— 10, p. 45.)
and /. A LMN ||| A DEF.
186
THEOKETICAL GEOMETRY
BOOK III
109.— Exercises
1. Prove the following construction for inscribing a A
similar to a given A DEF in
the circle LMN.
Draw a tangent HLG. Make
Z GLN = Z E, and Z HLM =
Z F. Join MN.
2. Inscribe an equilateral A
in a given circle.
3. Inscribe a square in a given circle.
4. Inscribe a regular pentagon in a given circle. (Use
protractor).
5. Inscribe a regular hexagon in a given circle. (Without
protractor).
6. Inscribe a regular • octagon in a given circle.
7. Two As LMN, DEF, each similar to a given A GHK,
are inscribed in a given circle. Prove A LMN EE A DEF.
8. In a given circle inscribe a A having its sides || to
the sides of a given A.
/
PROBLEM 7
To find the locus of the centres of circles
touching two given intersecting straight lines.
ABC, DBE be the two st. lines.
CONSTRUCTIONS 187
Construction. — Draw the bisectors FBG, HBK of the
^s made by AC and DE.
These bisectors make up the required locus.
Proof. — Take a point P in either FG or HK, and
draw PM _L AC, PN _L DE.
C L PBM = L PBN,
In As PMB, PNB, \ L PMB = L PNB,
\ and PB is common,
/. PM = PN. (1—14, p. 54.)
Hence, a circle described with centre P and radius
PM will pass through N.
".' L s at M, N are rt. z_ s,
.*. AC, DE are tangents to the circle.
(HI— 14, Cor. 1, p. 171.)
110. Definitions. — When a circle is within a triangle,
and the three sides of the triangle are tangents to the
circle, the circle is said to be inscribed in the triangle,
and is called the inscribed circle of the triangle.
When a circle lies without a triangle, and touches
one side and the other two sides produced, the circle
is called an escribed circle of the triangle.
188 THEORETICAL GEOMETRY BOOK lit
PROBLEM 8
To inscribe a circle in a given triangle.
B D C
Let ABC be the given A.
Bisect Z_s B and C and produce the bisectors to
meet at I.
Draw ID, IE, IF, _L BC, CA, AB respectively.
C L IBD = L IBF,
In As BID, BIF, L IDB = L IFB,
[ IB is common,
/. ID - IF. (1—14, p. 54.)
Similarly, ID = IE.
.". a circle described with centre I and radius ID
will pass through E and F.
And '.' the Ls at D, E and F are rt. Z_s,
/. the circle will touch BC, CA and AB.
(Ill— 14, Cor. 1, p. 171.)
CONSTRUCTIONS 189
PROBLEM 9
To draw an escribed circle of a given triangle.
A
\H
Let ABC be a given A having AB, AC produced to
G, H.
It is required to describe a circle touching the side
BC and the two sides AB, AC produced.
Bisect LS GBC, HCB and let the bisectors meet at
L. Draw J_s LP, LQ, LR to BC, CH, BG respectively.
f L PBL = L RBL,
In AS LBP, LBR, \ L LPB = L LRB,
[ LB is common,
.-. LP = LR. (1—14, p. 54.)
Similarly LP = LQ.
.*. a circle described with centre L and radius LP
will pass through R and Q.
V the L. s at P, Q and R are rt. L s,
.'. the circle will touch BC, and CA and AB pro-
duced. (Ill— 14, Cor. 1, p. 171.)
190
THEORETICAL GEOMETRY
BOOK III
PROBLEM 10
To describe a circle to touch three given straight
lines.
(a) If two of the lines are [| to each other, and
the third cuts them, two circles may be drawn to
touch the three lines.
N
Let ABC, DEF and GBEH be the three lines of which
AC II DF.
Bisect LS ABE, BED, and produce the bisectors to
meet at I.
Draw IL, IM, IN _i_ DE, EB, BA respectively.
As in problems 8 and 9 it may be shown that a
circle described with centre I and radius IL will touch
DE, EB and BA.
Similarly, a circle may be described on the other
side of BE to touch the three given st. lines.
(6) If the lines intersect each other forming a A,
four circles may be drawn to touch the three lines.
, EXERCISES
Let ABC be the A formed by the lines.
191
Draw the inscribed circle and the three escribed
circles of A ABC.
These four circles touch the three given st. lines.
ill.— Exercises
1. Make an L YXZ = 45°. Find a point P such that its
distance from XY is 3 cm., and its distance from XZ is 4 cm.
2. Make an L YXZ = 60°. Find a point P such that its
distance from XY is 4 cm., and its distance from XZ is 5 cm.
3. The bisectors of the Ls of a A are concurrent.
4. The bisectors of the exterior L. s at two vertices of a
A and the bisector of the interior L at the third vertex
are concurrent.
5. If a, b, c represent the numerical measures of the
sides BC, CA, AB respectively of A ABC, and 8 = J
(a + b + c),
192 THEORETICAL GEOMETRY BOOK III
(a) AF = s - a, BD = s - b, CE = s - c, when D, E
and F are the points of contact of BC, CA, AB with the
inscribed circle. (Diagram of Problem 8.)
(b) AR = s, BP = s-c, CP = s - b, where R and P
are the points of contact of AB produced and of BC with
an escribed circle. (Diagram of Problem 9.)
(c) If r be the radius of the inscribed circle, rs = the
area of A ABC.
(d) If fj be the radius of the escribed circle touching
BC, rl (s - a) = the area of A ABC.
6. If the base and vertical L of a A be given, find the
locus of the inscribed centre.
7. If the base and vertical L of a A be given, find the
loci of the escribed centres.
8. L, M, N are the centres of the escribed circles of
A PQR. Show that the sides of A LMN pass through the
vertices of A PQR.
9. If the centres of the escribed circles be joined, and
the points of contact of the inscribed circle with the sides
be joined, the As thus formed are similar.
10. Construct a A having given the base, the vertical
L and the radius of the inscribed circle.
11. Describe a circle cutting off three equal chords of
given length from the sides of a given A.
12. An escribed circle of A ABC touches BC at D and
also touches AB and AC produced. The inscribed circle
touches BC at E. Show that DE equals the difference of
AB and AC.
13. Circumscribe a square about a given circle.
14. Inscribe a circle in a given square.
15 Circumscribe a circle about a given square.
CONSTRUCTIONS
193
PROBLEM 11
About a given circle to circumscribe a triangle
similar to a given triangle.
K
L AM
Let ABC be the given circle and DEF the given A.
Construction. — Produce EF to G and H.
Draw any radius OA of the circle, and at O make
L AOB = L DFH, and L AOC = L DEG ; and produce
the arms to cut the circle at B, C.
At A, B, C draw tangents to the circle meeting at
K, L and M.
KLM is the required A.
Proof. — v /-s MAO and MBO in the quadrilateral
MBOA are rt. L. s,
.'. L M + L AOB = 2 rt. L. s.
= L DFE+^- DFH.
But, L AOB = L DFH,
/. L M = L DFE.
Similarly, L. L = L DEF.
/. L L+ L M = L DEF+ L DFE,
and /. L K = L EOF. (I — 10, p. 45.)
.-. A KLM HI A DEF.
194
THEORETICAL GEOMETRY
BOOK III
112.— Exercises
1. About a given circle circumscribe an equilateral A-
2. If two similar As be circumscribed about the same
circle, the As are congruent.
3. Describe a A LMN similar to a given A and such
that a given circle is touched by MN and by LM and LN
produced.
PROBLEM 12
To inscribe a circle in a given regular polygon.
B
C H D
Let AB, BC, CD, DE be four consecutive sides of a
given regular polygon.
It is required to inscribe a circle in the polygon.
Bisect L s BCD, CDE and produce the bisectors to
meet at O. Join OB. From O draw J_s OF, OG, OH,
OK to AB, BC, CD, DE respectively.
{BC = CD,
CO is common,
L OCB = L OCD,
/. L OBC = L ODC. (1—2, p. 16.)
But L ODC = \ L CDE and L ABC = L CDE,
.. L OBC = \ L ABC.
CONSTRUCTIONS
195
In the same manner it may be shown that if O be
joined to all the vertices of the polygon the joining
lines will bisect the Zs at the vertices.
C L OCG = L OCH,
In As OCG, OCH, I *- OGC = L OHC,
\ OC is common,
.-. OG = OH. (1—14, p. 54.)
In the same manner it may be shown that the J_s
from O to all of the sides are equal to each other,
and as the Ls at F, G, H, etc., are rt. Zs, a circle
described with O as centre and OF as radius will
sides and be inscribed in the polygon.
Au*z eui&b w*A***4 oV
196
THEORETICAL GEOMETRY
BOOK III
CONTACT OF CIRCLES
113. Definition. — If two circles meet each other
at one and only one point, they are said to touch
each other at that point.
THEOREM 16
If two circles touch each other, the straight line
joining their centres passes through the point of
contact.
Let two circles DEF, GEF,w of which the centres are
H, K respectively, cut each other at E, F.
Join HE, HF, KE, KF.
v HEF, KEF are isosceles As on the same base EF,
.*. HK is an axis of symmetry of the quadrilateral
HEKF and E, F are corresponding points. (I — 5, p. 24.)
.'. HK. bisects EF.
Let the circle GEF move so that the points Ef F
approach each other and finally coincide.
CONTACT OF CIRCLES
197
V L is the middle point of EF,
/. L coincides with E and F, the circles touch at
L, and the st. line HK passes through the point of
contact L.
Cor. i.— The straight line drawn from the point
of contact perpendicular to the line of centres is a
common tangent to the two circles.
Definition. — If two circles which touch each other
are on opposite sides of the common tangent at their
point of contact, and consequently each circle outside
the other, they are said to touch externally; if they
are on the same side of the common tangent, and con-
sequently one within the other, they are said to touch
internally.
Cor. 2.— If two circles touch externally, the
distance between their centres is equal to the sum
of their radii ; and conversely.
Cor. 3.— If two circles touch internally, the
distance between their centres is equal to the
difference of their radii; and conversely.
198 THEORETICAL GEOMETRY BOOK III
114.— Exercises
1. If the st. line joining the centres of two circles pass
through a point common to the two circumferences, the
circles touch each other at that point.
2. Find the locus of the centres of all circles which touch
a given circle at a given point.
^ 3. Draw three circles with radii 23, 32 and 43 mm. each
of which touches the other two externally.
w.**4. Draw a circle of radius 9 cm., and within it draw two
circles of radii 3 cm. and 4 cm., to touch each other exter-
ynally, and each of which touches the first circle internally.
5. Draw a circle of radius 85 mm., and within it draw two
circles of radii 25 mm. and 35 mm., to touch each other exter-
nally, and each of which touches the first circle internally.
6. Draw a A ABC with sides 5, 12 and 13 cm. Draw
three circles, with centres A, B and C respectively, each
of which touches the other two externally.
7. Construct the A ABC, having a = 5 cm., 6 = 4 cm.,
and c = 3 cm. Draw three circles with centres A, B and
C respectively, such that the circles with centres B and C
touch externally, and each touches the circle with centre A
internally.
8. Mark two points P and Q 10 cm. apart. With
centres P and Q, and radii 4 cm. and 3 cm., describe two
circles. Draw a circle of radius 5 cm. which touches each
of the first two circles externally. Find the distarice of
the centre from PQ.
« 9. Describe a circle to pass through a given point, and
touch a given circle at a given point.
. 10. If two circles touch each other, any st. line drawn
through the point of contact will cut off segments that
contain equal La.
EXERCISES 199
/ 11. Two circles AGO, BDO touch, and through O, st.
lines AOB, COD are drawn. Show that AC || BD.
12. If two || diameters be drawn in two circles which
touch one another, the point of contact and an extremity
of each diameter are in the same st. line.
13. Describe a circle which shall touch a given circle,
have its centre in a given st. line, and pass through a
given point in the st. line.
14. Describe three circles having their centres at three
given points and touching each other in pairs. Show that
there are four solutions.
15. Two circles touch a given st. line at two given points,
and also touch each other; find the locus of their point of
contact.
16. If through the point of contact of two touching
circles a st. line be drawn cutting the circles again at two
points, the radii drawn to these points are ||.
» 17. In a given semi-circle inscribe a circle having its
radius equal to a given st. line.
* 18. Inscribe a circle in a given sector.
19. A circle of 2-5 cm. radius has its centre at a dis-
tance of 5 cin. from a given st. line. Describe four circles
each of 4 cm. radius to touch both the circle and the st.
line.
20. If DE be drawn || to the base GH of a A FGH
to meet FG, FH at D, E respectively, the circles described
about the As FGH, FDE touch each other at F.
21. Two circles with centres P, Q touch externally and
a third circle is drawn, with centre R, which both the
first circles touch internally. Prove that the perimeter of
A PQR = the diameter of the circle with centre R.
200 THEORETICAL GEOMETRY . BOOK III
t
Miscellaneous Exercises
1. If two chords of a circle intersect at rt. L s, the
sum of the squares on their segments is equal to the
square on the diameter.
2. Find a point in the circumference of a given circle,
the sum of the squares on whose distances from two given
points may be a maximum or minimum.
3. AOB, COD are chords cutting at a point O within
the circle. Show that L BOG equals an L at the circum-
ference, subtended by an arc which is equal to the sum of
the arcs subtending LS BOG, AOD.
4. Two chords AB, CD intersect at a point O without
a circle. Show that L AOC equals an L at the circum-
ference subtended by an arc which is equal to the
difference of the two arcs BD, AC intercepted between
OBA and ODC.
5. Two circles touch externally at E, and are cut by a
st. line at A, B, C, D. Show that L AED is supplementary
to L BEC.
6. If at a point of intersection of two circles the
tangents drawn to the circles be at rt. L s, the st. line
joining the points where these tangents meet the circles
again, passes through the other point of intersection of the
circles.
7. Find a point within a given A at which the three
sides subtend equal LS. When is the solution possible?
8. Through one of the points of intersection of two
given circles draw the greatest possible st. line terminated
in the two circumferences.
9. Through one of the points of intersection of two
given circles draw a st. line terminated in the two circum-
ferences and equal to a given st. line.
MISCELLANEOUS EXERCISES 201
10. Describe a circle of given radius to touch two given
circles.
11. DEF is a st. line cutting BC, CA, AB, the sides of
A ABC, at D, E, F respectively. Show that the circles
circumscribed about the As AEF, BFD, CDE, ABC, all
pass through one point.
12. Two circles touch each other at A and BAG is
drawn terminated in the circumferences at B, C. Show
that the tangents at B, C are ||.
13. D, E, F are any points on the sides BC, CA, AB of
A ABC. Show that the circles circumscribed about the
As AFE, BDF, CED pass through a common point.
14. Two arcs stand on a common chord AB. P is any
point on one arc and PA, PB cut the other arc at C, D.
Show that the length of CD is constant.
15. ACB is an L in a segment. The tangent at A is ||
to the bisector of L ACB and meets BC produced at D.
Show that AD = AB.
16. Describe a circle of given radius to touch two given
intersecting st. lines.
17. In the A ABC, the bisector of L A meets BC at D.
O is the centre of a circle which touches AB at A and
passes through D. Prove that CD _L AC.
18. The st. line BC of given length moves so that B and
C are respectively on two given fixed st. lines AX and AY.
Prove that the circumcentre of A ABC lies on the
circumference of a circle with centre A.
19. ABC is an isosceles A in which AB = AC. D is
any point in BC. Show that the centre of the circle
ABD is the same distance from AB that the centre of the
circle ACD is from AC.
202 THEORETICAL GEOMETRY , BOOK III
20. E, F, G, H are the points of contact of the sides of
a quadrilateral A BCD circumscribed about a circle. Prove
that the difference of two opposite LS of A BCD = twice
the difference of two adjacent L s of EFGH.
21. ABC is a A in which AX, BY are _L_ BC, CA re-
spectively. Prove that the tangent at X to the circle CXY
passes through the middle point of AB ; and the tangent
at C to the same circle || AB.
22. The inscribed circle of A ABC touches BC at D.
Prove that the circles inscribed in As BAD, CAD touch
each other.
23. O is the circum centre of the A ABC, and AO, BO,
CO produced meet the circumference in D, E, F. Prove
A DEF = A ABC.
24. ABC is a rt.- L d A, A being the rt. L . Prove that
BC = the difference between the radius of the inscribed
circle and the radius of the circle which touches BC and
the other two sides produced.
25. Describe two circles to touch two given circles, the
point of contact with one of these given circles being
given.
26. Circles through two fixed points A and B intersect
fixed st. lines, which terminate at A and are equally
inclined to AB on opposite sides of it, in the points L, M.
Prove that AL + AM is constant.
27. AB is a diameter and CD a chord of a given circle.
AX and BY are both ± CD. Prove that CX = DY.
28. Through a fixed point A on a circle any chord AB is
drawn and produced to C making BC = AB. Find the
locus of C.
MISCELLANEOUS EXERCISES 203
29. Construct a A having given the base, the vertical /-,
and the length of the median drawn from one end of the
30. If the sum of one pair of opposite sides of a quadri-
lateral is equal to the sum of the other pair, a circle may
be inscribed in the quadrilateral.
31. Construct a A having given the vertical /_, the
base, and the point where the bisector of the vertical L
cuts the base.
32. From the ends of a diameter BC of a circle, ||
chords BE, CF are drawn, meeting the circle again in E
and F. Prove that EF is a diameter.
33. ACFB and ADEB are fixed circles; CAD, CBE and
DBF are st. lines. Prove that CF and DE meet at a
constant £_•
34. A, B, C, D are four points in order on the circum-
ference of a circle, and the arc AB = the arc CD. If AC
and BD cut at E, the chord which bisects z_s AEB,
CED is itself bisected at E.
35. AB, AC are tangents at B, C to a circle, and D is
the middle point of the minor arc BC. Prove that D is
the centre of the inscribed circle of the A ABC.
36. Construct an equilateral A whose side is of given
length so that its vertices may be on the sides of a given
equilateral A.
37. D, E, F are the points of contact of the sides BC,
CA, A B of a A ABC with its inscribed circle. FK is
_L DE, and EH is J_ FD. Prove HK II BC.
38. Tangents are drawn from a given point to a system
of concentric circles. Find the locus of their points of
contact.
204 THEORETICAL GEOMETRY , BOOK III
39. From a given point A without a given circle draw
a secant ABC such that AB = BC.
40. EF is a fixed chord of a given circle, P any point
on its circumference. EM J_ FP and FN _L EP. Find the
locus of the middle point of MN.
41. K is the middle point of a chord PQ in a circle
of which O is the centre. LKM is a chord. Tangents
at L, M meet PQ produced at G, H respectively. Prove
A OGL = A OHM.
42. LM is the diameter of the semi-circle LNM in which
arc LN > arc NM, and ND J_ LM. A circle inscribed in
the figure bounded by ND, DM and the arc NM touches
DM at E. Show that LE = LN ; and hence give a con-
struction for inscribing the circle.
43. GK is a diameter and O the centre of a circle. A
tangent KD = KO. From O a J_ OE is drawn to GD. KE
is joined and produced to meet the circumference in F.
Prove that FE = FG.
44. LPM and LQRM are two given segments on the
same chord LM. If P moves on the arc LPM such that
LQP and MRP are st. lines, the length of QR is constant.
45. EFP, EFRS are two circles and PFR, PES are st.
lines. O is the centre of the circle EFP. Prove that
PO _L RS.
46. E, F are fixed points on the circles EPD, FQD, and
PDQ is a variable st. line. PE, QF intersect at R. Find
the locus of R.
47. The circle PEGF passes through the centre G of the
circle QEF, and P, E, Q are in a st. line. Prove that
PQ = PF.
48. Through two points on a diameter equally distant
from the centre of a circle, || chords are drawn, show that
MISCELLANEOUS EXERCISES 205
these chords are the opposite sides of a rectangle inscribed
in the circle.
49. If through the points of intersection of two circles
any two || st. lines be drawn and the ends joined towards
the same parts, the figure so formed is a ||gm.
50. Any two || tangents are drawn, one to each of two
given circles ; a st. line is drawn through the points of
contact, show that the tangents to the circles at the other
points of intersection are also ||.
51. The hypotenuse of a rt.-/_d A is fixed and the other
two sides are moveable, find the locus of the point of
intersection of the bisectors of the acute /_s of the A.
52. From the middle point L of the arc MLN of a circle
two chords are drawn cutting the chord MN and the cir-
cumference. Show that the four points of intersection are
concyclic.
53. If from one end of a diameter of a circle, two st.
lines be drawn to the tangent at the other end of the
diameter, the four points of intersection — with the circle,
and with the tangent — are concyclic.
54. ABC is a diameter of a circle, B being the centre.
AD is a chord, and BE JL to AC cutting the chord at E.
Show that BCDE is a cyclic quadrilateral; and that the
circles described about ABE and the quadrilateral BCDE,
are equal.
55. Two circles intersect at A and B. From A two
chords AC and AE are drawn one in each circle making
equal z_s with AB, st. lines CBD and EBF are drawn to
cut the circles at D and F, prove C, F, D, E concyclic;
also prove As FCA and DEA similar.
56. ABC is a A and any circle is drawn passing through
B, and cutting BC at D and AB at F; another circle is
206 THEORETICAL GEOMETRY ' BOOK Ilf
drawn passing through C and D and intersecting the
former circle at E and AC at G. Prove A, F, E, G are
concyclic.
57. If two equal circles intersect, the four tangents at
the points of intersection form a rhombus.
58. If two equal circles cut, and at G, one of the points
of intersection, chords be drawn in each circle, to touch
the other circle, these chords are equal.
59. Two equal circles, centres O and P, touch externally
at S, SQ and SR are drawn J_ to each other cutting the
circumferences at Q and R respectively. Show that O, P, Q
and R are the vertices of a ||gm.
60. AB, CD, and EF are || chords in a circle, prove that
the As ACE and BDF are congruent; also ACF and BDE;
also ADF and BCE.
61. On the circumference of a circle are two fixed points
which are joined to a moveable point either inside or
outside the circle. If these lines intercept a constant arc,
find the locus of the point.
62. KL is any chord of a circle and H the middle point
of one of the arcs, any st. line HED cuts KL at E and
the circumference at D. Show that HL is a tangent to
the circle about LED, and HK a tangent to that about
KED.
63. Two circles intersect at E and F. From any point P
on the circumference of one of them st. lines PE and PF
are drawn to meet the circumference of the other at Q and
R, show that the length of the straight line QR is constant.
[Take P both on the major arc and on the minor arc.]
64. HKL is a A having L H acute; on KL as diameter
a circle, centre O, is described cutting HK at D and HL
at E. Show that L ODE = L H.
MISCELLANEOUS EXERCISES 207
65. P is a point external to two concentric circles whose
centre is O, PQ is a tangent to the outer circle and PR
and PS are tangents to the inner circle. Show that L
RQS is bisected by QO.
66. If the extremities of two || diameters in two circles
be joined by a st. line which cuts the circles, the tangents
at the points of intersection are ||. Show, that this is true
for the four cases that arise.
67. KLMN is a ||gm, through L and N two || st. lines
are drawn cutting MN at F and KL at E, show that the
circles described about the As KNE and LMF are equal.
68. EFGH is a quadrilateral having EF || HG and EH = FG.
From E a st. line EK is drawn || FG meeting HG at K.
Show that circles described about the As EHG, EKG are
equal.
69. From any point P on the circumference of a circle
PD, PE and PF are perpendiculars to a chord QR, and to
the tangents QT and RT. Show that the As PED and
PFD are similar.
70. A quadrilateral having two || sides is described about
a circle. Show that the st. line drawn through the centre
|| to the || sides and terminated by the nonparallel sides is
one quarter of the perimeter of the quadrilateral.
71. CD is a diameter of a circle centre O; chords CF
and DG intersect within the circle at E. Show that OF
is a tangent to the circle passing through F, G and E.
72. EF is a chord of a circle and EP a tangent; a st.
line PG || to EF meets the circle at G ; prove that the
As EFG and EPG are similar.
73. The diagonals of a quadrilateral are JL ; show that
the st. lines joining the feet of the perpendiculars from
208 THEORETICAL GEOMETRY %• BOOK III
the intersection of the diagonals on the sides form a cyclic
quadrilateral.
74. Two chords of a circle intersect at rt. /_s and
tangents are drawn to the circle from the extremities of
the chords \ show that the resulting quadrilateral is cyclic.
75. A quadrilateral is described about a circle and its
vertices are joined to the centre cutting the circumference
in four points. Show that the diagonals of the quadri-
lateral formed by joining these four points are J_.
76. DEF is a A inscribed in a circle whose centre is O.
On EF any arc of a circle is described and ED, FD, or
these lines produced, meet the arc at P, Q. Show that
OD, or OD produced, cuts PQ at rt. /_s.
77. PQRS is a ||gm and the diagonals intersect at E.
Show that the circles described about PES and QER touch
each other; and likewise those about PEQ and RES.
78. Two equal circles intersect at E and F ; with centre
E and radius EF a circle is described cutting the circles at
G and H. Show that FG and FH are tangents to the
equal circles.
79. If from any point on the circumference of a circle
perpendiculars be drawn to two fixed diameters, the line
joining their feet is of constant length.
80. From the extremities of the diameter of a circle
perpendiculars are drawn to any chord. Show that the
centre is equally distant from the feet of the perpendiculars.
81. EF and GH are || chords in a circle, F and H being
towards the same parts ; a point K is taken on the
circumference such that GF bisects L HGK. Prove GK = EF.
82. Two circles intersect at D and E, and KEL and PEQ
are two chords terminated by the circumferences. Show that
the As DKP and DLQ are similar.
MISCELLANEOUS EXERCISES 209
83. If from two points outside a circle, equally distant
from the centre and situated on a diameter produced,
tangents be drawn to the circle, the resulting quadrilateral
is a rhombus.
84. If the arcs cut off by the sides of a quadrilateral
inscribed in a circle be bisected and the opposite points be
joined, these two lines shall be J_. (Note. — Use Ex. 3.)
85. PQ is a fixed st. line and PM, QN are any two
|| st. lines, M and N being towards the same parts. The
LS MPQ and NQP are bisected by PR and QR. Find the
locus of R.
86. If the /_s of a A inscribed in a circle be bisected
by lines which meet the circumference, and a new A be
formed by joining these points on the circumference, its
sides shall be _L to the bisectors.
87. If two circles touch each other internally, and a st.
line be drawn || to the tangent at the point of contact,
the two intercepts between the circumferences subtend
equal /_s at the point of contact.
88. ABC is a A inscribed in a circle and BA is pro-
duced to E; D is any point in AE; circles are described
through B, C, D and through B, C, E ; CFDG cuts the
circles ABC, EBC in F and G. Prove that As ADF and
DEG are similar.
89. Draw a tangent to a circle which shall bisect a given
||gm which is outside the circle.
90. In a given circle draw a chord of fixed length which
shall be bisected by a given chord.
91. In a given circle draw a chord which shall pass
through a given point and be bisected by a given chord.
How many such chords can be drawn?
210 THEORETICAL GEOMETRY < BOOK III
92. Describe a circle with given radius to touch a given
st. line and have its centre in another given st. line.
93. Describe a circle of given radius to pass through a
given point and touch a given st. line.
94. Describe a circle to touch a given circle at a given
point and a given st. line.
95. In a given st. line find a point such that the st.
lines joining it to two given points may be (a) JLs, (6)
make a given /_ with each other.
96. Describe a circle of given radius to touch a given
circle and a given st. line.
97. Describe a circle to touch a given circle and a given
st. line at a given point.
98. Inscribe in a given circle a A one of whose sides
shall be equal to a given st. line, and such that the other
two may pass through two given points respectively.
99. Place a chord PQ in a circle so that it will pass
through a given point O within the circle, and such that
the difference between OP and OQ may be equal to a
given st. line.
100. Find two points on the circumference of a given
circle which shall be concyclic with two given points P
and Q outside the circle.
101. Describe a square (EFGH) having given the point F
and two points P and Q in the sides FE and EH
respectively.
102. Describe a square (EFGH) having given the point G
and two points P and Q in the sides FE and EH
respectively.
103. Describe a square so that its sides shall pass
respectively through four given points.
MISCELLANEOUS EXERCISES 211
104. If three circles touch externally at P, Q, R and PQ
and PR meet the circumference of QR at D and E, then
DE is a diameter, and is || to the line joining the centres
of the other two circles.
105. Two equal circles intersect so that the tangents at
one of the points of intersection are J_s. Show that the
square on the diameter is twice the square on the common
chord.
106. LMN is a rt.-^-d A, L being the rt. L, and LD is _L
to MN. Show that LM is a tangent to the circle LDN.
107. PQ is a tangent to a circle and PRS a secant
passing through the centre, QN is _L to' PS. Show that
QR bisects L PQN.
108. LMN is a A inscribed in a circle whose centre is
O. Show that the radius OL makes the same L with
LM that the -L from L to MN makes with LN.
109. If two chords of a circle be J_, the sum of one
pair of opposite intercepted arcs is equal to the sum of the
other pair.
110. On the sides of a quadrilateral as diameters circles
are described. Show that the common chords of every
adjacent pair of circles is || to the common chord of the
remaining pair.
111. Two equal circles are so situated that the distance
between their nearest points is less than the diameter of
either circle. Show how to draw a st. line cutting them so
as to be trisected by the circumferences.
112. LMN is a A and D, E, F are the middle points of
MN, NL and LM respectively; if LP is the perpendicular
on MN, show that D, P, E, F are coney clic.
212 THEORETICAL GEOMETRY < BOOK III
113. QR is a fixed chord of a circle and P a moveable
point on the circumference. Find the locus of the inter-
section of the diagonals of the ||gm having PQ and QR
for adjacent sides.
114. If a quadrilateral having two || sides is inscribed in
a circle, show that the four perpendiculars from the middle
point of an arc cut off by one of the || sides, to the two
diagonals and to the nonparallel sides, are equal.
115. ABCD and A'B'C'D' are any rectangles inscribed in
two concentric circles respectively. P is on the circum-
ference of the former circle and P' on the latter. Prove
PA'2 + PB'2 -f PC'2 + PD'2 = P'A2 -f P'B2 + P'C2 + P'D2.
116. A point Y is taken in a radius of a circle whose
centre is O ; on OY as base an isosceles A XOY is
described having X on the circumference; XO and XY
are produced to meet the circumference at D and Z
respectively, and E is the point between D and Z where
the perpendicular from O to OY cuts the circle. Show
that the arc DE is one-third of arc EZ.
BOOK IV
RATIO AND PROPORTION
115. Definitions. — The ratio of one magnitude to
another of the same kind is the number of times that
the first contains the second; or it is the part, or
fraction, that the first magnitude is of the second.
Thus the ratio of one magnitude to another is the
same as the measure of the first when the second is
taken as the unit.
If a st. line is 5 cm. in length, the ratio of its
length to the length of one centimetre is 5, that is,
the st. line is to one centimetre as 5 is to 1.
If two st. lines A, B are respectively 8 inches and
3 inches in length, then the ratio of A to B is 8 to 3.
The ratio of one magnitude A to another B is
written either s» or A ; B.
B
When the form — is used, the upper magnitude is
called the numerator, and the lower the denominator;
and when the form A : B is used, the first magnitude
is called the antecedent, and the second the conse-
quent. The two magnitudes are called the terms of
the ratio.
116. Definitions. — Proportion is the equality of
ratios, i.e., when two ratios are equal to each other,
the four magnitudes are said to be in proportion.
The equality of the ratios of K to L and of M to
N may be written in any one of the three forms: —
- = |£« K : L = M ; N or K ; L ; ; M : N ; and is read
11 K is to L as M is to N."
213
214 THEORETICAL GEOMETRY BOOK IV
The four magnitudes in a proportion are called
proportionals.
The first and last are called the extremes, and the
second and third are called the means.
The first two magnitudes of a proportion must be
of the same kind, and the last two must be of the
same kind ; but the first two need not be of the
same kind as the last two. Thus in the proportion
^ = — • D and E may be lengths of lines, while F and
H are areas.
117. Definitions. — Three magnitudes are said to be
in continued proportion, or in geometric progres-
sion, when the ratio of the first to the second equals
the ratio of the second to the third.
Three magnitudes L, M, N, of the same kind, are in
continued proportion, if ^ = ^-
e. g. : — L = 4 cm., M = 6 cm., N = 9 cm.
The second magnitude of a continued proportion is
called the mean proportional, or geometric mean, of
the other two.
118. Two magnitudes of the same kind are com-
mensurable when each contains some common measure
an integral number of times.
Two magnitudes of the same kind are incommen-
surable when there is no common measure, however
small, contained in each of them an integral number
of times.
RATIO AND PROPORTION 215
The diagonal and side of a square are incommen-
surable ; the ratio of the diagonal to the side being
V* : 1-
The side of an equilateral triangle and the per-
pendicular from a vertex to the opposite side
are incommensurable; the ratio of a side to the
perpendicular being 2 ; |/3.
l/2~= 1-414 nearly, and y/3~= 1732 nearly, but
while these roots may be calculated to any required
degree of accuracy they cannot be exactly found.
Thus there is no straight line however short that is
O
contained an integral number of times in both the
diagonal and side of a square; or in both the side
and altitude of an equilateral triangle.
The treatment of incommensurable magnitudes is
too difficult for an elementary text-book, but as in
algebra, the relations that are obtained in geometry
for commensurable magnitudes hold good also for
incommensurable magnitudes.
119. The following simple algebraic theorems are
used in geometry : —
1. If £ = f , ad = fee.
b d
If four numbers be in proportion, the product of
the extremes is equal to the product of the means.
9 Tf a c a =- ^
— . -Ll — — — , — — .
o d c d
If four numbers be in proportion, the first is to
the third as the second is to the fourth.
t
216 THEORETICAL GEOMETRY BOOK IV
When a proportion is changed in this way the
second proportion is said to be formed from the first
by alternation.
In order that a given proportion may be changed
by alternation, the four magnitudes must be of the
same kind.
2 ft. 4 ft. 2 ft.
e.g.: — — -^ = _ and, by alternation, — — =
5 ft. 10 ft. 4 ft.
*Jk; but from the proportion st line D = area F
10 ft. st. line E area G
another proportion cannot be inferred by alternation.
3f£ a c b d
.11 — — -, - = -.
b d a c
If four numbers be in proportion, the second is
to the first as the fourth is to the third.
When a proportion is changed in this way the
second proportion is said to be formed from the first
by inversion.
4 jf a _. c a + b __ c + d
b~ d' ~~b~ d
If four numbers be in proportion, the sum of the
first and second is to the second as the sum of the
third and fourth is to the fourth.
T«a_ c a — b _ c — d
b ~ d' ~T~ ~d~
If four numbers be in proportion, the difference
of the first and second is to the second as the
difference of the third and fourth is to the fourth.
a c a + b c + d
a
b.
-7 -v - =- = - — ,
b d a — b c — d
EATIO AND PROPORTION 217
If four numbers be in proportion, the sum of the
first and second terms is to the difference of the
first and second terms as the sum of the third and
fourth terms is to the difference of the third and
fourth terms.
7. If ? = £ = * etc., then each of the equal
0 CL T
e ,. a + c + e + etc.
fractions = — — —
b + d + / + etc.
If any number of ratios, the terms of which are
all magnitudes of the same kind, be equal to each
other, the sum of the numerators divided by the
sum of the denominators equals each of the given
ratios.
. -r/. 7 -id c j ci b
8. If ad = be, r = T, and - = _
b d c d
If the product of two numbers be equal to the
product of two other numbers, one factor of the
first product is to a factor of the second product
as the remaining factor of the second is to the
remaining factor of the first.
120. If a given straight
line MN be divided internally tf - — 1 ^
at a point P, the internal
segments PM, PN are the distances from P to the
ends of the given straight line.
Similarly, if a point P be taken in a given straight
line MN produced, the dis-
i~l J- — p tances from P to the ends of
the given straight line, PM,
PN, are called the external segments of the straight
218 THEORETICAL GEOMETRY BOOK IV
line, or the given straight line is said to be divided
externally at the point p.
*21< There is only one
1 point where a straight line
MN is divided internally into
segments IN/IP, PN that have a given ratio — .
For, if possible, let it be divided internally at
P and Q such that -p^ and -^ each equals
Then W£=m
. MP+ PN = MQ+ QN.
PN QN
MN __ MN
/l'6'' PN ~~ Ql\r
PN = QN.
and .*. Q coincides with p.
Similarly, there is only one point where a straight
line MN is divided externally
into segments MP, PN that „ 1 1 —
a N Q
have a given ratio - .
For, if possible, let it be divided externally at
P and Q such that ^p and Q^- each equals - -
m, MP MQ
Then ^rr = ^ •
MP - PN MQ - QN
PN QN
MN MN
*.e., PN ~ QN "
PN = QN.
and .'. Q coincides with P.
(5, § 119.)
EXERCISES
THEOREM 1
219
Triangles of the same altitude are to each other
as their bases.
B
Hypothesis. — In As ABC, DEF; AX _L BC, DY _L EF
and AX = DY.
To prove that
A ABC BC
A DEF " EF
Construction. — On BC and EF construct the rect-
angles HC and LF, having HB = AX and LE = DY.
Proof. — Let BC and EF contain a and b units of
length respectively, and AX or DY contain c units.
A ABC - £ HB.BC = 4 ca. (II — 4, p. 100.)
A DEF = i LE.EF = \ cb.
• A ABC _ \ ca a _ BC
* A DEF ~ 1'cb = I = EF '
122.— Exercises
1. As on equal bases are to each other as their altitudes.
2. If two As are to each other as their bases, their
altitudes must be equal.
3. 1 1 gins of equal altitudes are to each other as their bases.
4. Construct a A equal to £ of a given A.
5. Construct a ||gm equal to f of a given ||gm.
220 THEORETICAL GEOMETRY BOOK IV
6. ABC, DEF are two As having AB = DE and L B =^
L E. Show that A ABC : A DEF = BC : EF.
7. The rectangle contained by two st. lines is a mean
proportional between the squares on the lines.
8. If two equal As be on opposite sides of the same
base, the st. line joining their vertices is bisected by the
common base, or the base produced.
9. The sum of the J_s from any point in the base of an
isosceles A to the two equal sides equals the J_ from
either end of the base to the opposite side.
10. The difference of the J_s from any point in the base
/produced of an isosceles A to the equal sides equals the
J_ from either end of the base to the opposite side.
11. The sum of the Xs from any point within an.
equilateral A to the three sides equals the J_ from any
vertex to the opposite side.
12. If st. lines AO, BO, CO are drawn from the vertices
of a A ABC to any point O- and AO, produced if necessary,
cuts BC at D,
AAOB BD
AAOC ~ DC'
13. In any A ABC, F is the middle point of AB, E is the
middle point of AC, and BE, CF intersect at O. Show that
AO produced bisects BC ; that is, the medians of a A are
concurrent.
14. ABC is a A and O is any point. AO, BO, CO,
produced if necessary cub BC, CA, AB at D, E, F respec-
tively, a1} a2, bl9 62, clf c2, are respectively the numerical
measures of BD, DC, CE, EA, AF, FB. Show that «x ^ cx
= a2 62 c2. (This is known as Ceva's Theorem.)
15. The four As into which a quadrilateral is divided
by its diagonals are proportional.
EXERCISES 221
16. DEF is a A ; G is a point in DE such that DG = 3
GE, and H is a point in DF such that FH = 3 HD. Show
that A FGH = 9 A EGH.
17. St. lines DG, EH, FK drawn from the vertices of
A DEF to meet the opposite sides at G, H, K pass through
a common point O. Prove that Q-Q 4- pTj ~h pi7 = ^.
18. In A DEF, G is taken in side EF such that EG =
2 GF, and H is taken in side FD such that FH = 2 HD.
DG and EH intersect at O. Prove that -\ pEf- = gj-
222
THEORETICAL GEOMETRY
THEOREM 2
BOOK IV
A straight line drawn parallel to the base of a
triangle cuts the sides, or the sides produced,
proportionally.
E* 7-D
Hypothesis. — In A ABC, DE || BC.
To prove that •—— = —-•
DA EA
Construction. — Join BE and DC.
Proof.— v DE || BC,
/. A BDE = A CDE
A BDE A CDE
" A ADE " A ADE"
(II— 5, p. 101.)
v As BDE, ADE have the same altitude, viz., the _|_
from E to AB,
In the same way,
A CDE = CE
A ADE ~ EA
BD
DA
CE
EA
EATIO AND PROPORTION
223
N.B. — By placing D on AB and E on AC* in all
three figures the proof applies to all.
Cor. — In the first figure,
BD+DA CE+EA 6 ^^
DA
CE
EA'
DA
AB
EA
AC
AE
X^ bV inverting.
Again,
BD CE
DA ~ EA'
DA
EA
CE
DA-fBD EA + CE
BD
AB
BD
BD
AB
CE
AC
CE~
CE
AC
by inverting,
by addition,
i
by inverting.
Similar proofs may be given for the second and
third figures.
Thus we see that where a line is parallel to the
base of a triangle we may form a proportion by
taking- the whole side or either of the segments,
in any order, for the terms of the first ratio, pro-
vided we take the corresponding parts of the other
side to form the terms of the other ratio in the
proportion.
224
THEORETICAL GEOMETRY
THEOREM 3
BOOK IV
(Converse of Theorem 2)
If two sides of a triangle, or two sides produced,
be divided proportionally, the straight line joining
the points of section is parallel to the base.
Hypothesis.— In A ABC, ^ = -^
To prove that DE || BC.
Construction. — Draw DF i BC,
Proof.—
TD J. BD
But
DF
BD
DA
BD
DA
CE
EA
BC,
CF
FA"
CE
EA"
CF
: FA'
to cut AC at F.
(IV— 2, p. 222.)
And .'. E coincides with F.
/. CE || BC.
123.— Exercises
1. The st. line drawn through the middle point of one
side of a A, and || to a second side bisects the third side.
2. The st. line joining the middle points of two sides of
a A is || to the third side.
^3. If two sides of a quadrilateral be ||, any st. line
drawn || to the || sides and cutting the other sides, will
cut these other sides proportionally.
\
EXERCISES 225
4. A BCD is a quadrilateral having AB || DC. P, Q are
points in AD, BC respectively such that AP : PD = BQ : QC.
Show that PQ || AB or DC.
5. If two st. lines are cut by a series of
il st. lines, the intercepts on one are proper-
tional to the corresponding intercepts on the
other.
6. D, E are points in AB, AC, the sides
A ABC, such that DE || BC ; BE, CD
meet at F. Show that AADF = A AEF.
Show also that AF bisects DE and BC.
7. Through D, any point in the side BC of A ABC, DE, DF
are drawn || AB, AC respectively and meeting AC, AB at
E, F. Show that A AEF is a mean proportional between
As FBD, EDC.
8. ACB, ADB are two As on the same base AB. E is
any point in AB. EF is || AC and meets BC at F. EG
is || AD and meets BD at G. Prove FG II CD.
9. D is a point in the side AB of A ABC ; DE is drawn
|| BC and meets AC at E ; EF is drawn II AB and meets
BOat F. Show that AD : DB = BF : FC.
^^) From a given point M in the side DE of A DEF,
draw a st. line to meet DF produced at N so that MN is
bisected by EF.
11. PQRS is a ||gm, and from the diagonal PR equal
lengths PK, RL are cut off. SK, SL when produced meet
PQ, RQ respectively at E, F. Prove EF || PR.
12. DEF is a A in which K, M are points in the side
DE and L, N are points in the side DF such that KL and
MN are both || EF. Find the locus of the intersection of
KN and LM.
13. O any point within a quadrilateral PQRS is joined
to the four vertices and in OP any point X is taken. XY
226 THEORETICAL GEOMETRY BOOK IV
(
is drawn || PQ to meet OQ at Y ; YZ is drawn || QR to meet
OR at Z; and ZV is drawn || RS to meet OS at V. Prove
that XV || PS.
14. O is a fixed point and P moves along a fixed st.
line. Q is a point in OP, or in OP produced in either
direction, such that OQ : QP is constant. Find the locus
of Q.
15. L is any point in the side DE of a A DEF. From
L a line drawn || EF meets DF at M. From F a line
drawn || ME meets DE produced ab N. Prove that
DL : DE = DE : DN.
16. If from the vertex of a A perpendiculars are drawn
to the bisectors of the exterior L. s at the base, the line
joining the feet of the perpendiculars is || the base.
PROBLEM 1
To divide a given straight line into any number
of equal parts.
(Alternative proof for I Prob. 8)
A K L M N B
\ "A
\ \
G
H
Let AB be the given straight line.
At A draw AC making any angle with AB and
from AC cut off in succession the required number of
equal parts. AD, DE, EF, FG, GH.
Join HB and through D, E, F, G draw lines || BH
cutting AB at K, L, M, N.
Then AK = KL - LM = MN = NB.
RATIO AND PROPORTION 227
In A AEL, DK || EL,
. AD _ AK /jY 9 „ 999 \
" DE-KL' ~2> P- A
But AD = DE, .. AK = KL,
In A AFM, EL || FM,
. AE _ AL
* EF LM*
But AE = 2 EF, :. AL = 2 LM
.-. LM = AK or KL.
In the same way it may be proved that AK = KL =
LM = MN = NB.
PROBLEM 2
To find a fourth proportional to three given
straight lines taken in a given order.
D G H E
L
Let A, B, C be the three given st. lines.
From a point D draw two st. lines DE, DF.
Cut off DG = A, GH = B, DK = C.
Join GK. Through H draw HL || GK meeting DF
in L.
Then KL is the required fourth proportional.
In A DHL, GK || HL
•• els - RC <IV-2> P- 222->
A C
'e'> B ~ KL
.". KL is the required fourth proportional.
228 THEORETICAL GEOMETRY BOOK IV
t
PROBLEM 3
To divide a given straight line in a given ratio.
A H B
<^
Let AB be the given si line, and - the given ratio.
Draw AE making any Z with AB.
On AE cut off AF = C, FG = D.
Join BG, and through F draw FH || GB.
In A ABG, V FH || GB,
" T^ = FG"- <IV-2,p.222.)
But AF = C, and FG = D,
AH C
' HB ~ D"
PROBLEM 4
To divide a given straight line similarly to a
given divided line.
C . |E |F D
Let AB be the given st. line, and CD the given
line divided at E and F.
At A draw AG making any angle with AB.
From AG cut off AH = CE, HK = EF, KL - FD. Join
BL. Through H, K draw HIM, KM both || BL.
EXERCISES 229
Then AB is divided at N and M similarly to CD.
Through H draw HPQ || AB.
Proof.— In A AMK, NH n MK,
- (IV-2)P.222.)
In A HQL, PK li QL,
. HP _ HK
* PQ KL"
But HP = NM and PQ = MB,
. NM_HK. j
* MB ~ KL W> P- b7''
AN _CE -, NM _ EF
•"• NM ~EF j MB"FD"
Both these relations are contained in
AN ^ NM = MB
CE ~ EF ~~ FD "
124.— Exercises
1. Divide the area of a given A into parts that are in
the ratio of two given st. lines.
2. Divide the area of a ||gm into parts that are in the
ratio of two given st. lines.
3. Find a third proportional to two given st. lines.
Show how two third proportionals, one greater than either
of the given st. lines and the other less than either, may
be found.
4. Divide a given st. line externally so that the ratio of
the segments may equal the ratio of two given st. lines.
5. BAG is a given L and P is a given point. Through
P draw a st. line DPE cutting AB at D and AC at E such
that DP : PE equals the ratio of two given st. lines.
230 THEORETICAL GEOMETRY BOOK IV
6. Divide a given st. line in the ratio 2 f 3 | 5.
7. Construct a A having its sides in the ratio 2 ! 3 ; 4,
and its perimeter equal to a given st. line.
8. From a given point P outside the L XOY draw a
line meeting OX at Q and OY at R so that PQ : QR =
a given ratio.
BISECTOR THEOREMS
THEOREM 4
If the vertical angle of a triangle is bisected by
a straight line which cuts the base, the segments
of the base are proportional to the other sides of
the triangle.
A ,-'
B- . D C
Hypothesis. — In A ABC, AD bisects L BAG.
2V,
Construction. — Through C draw CE || AD to meet BA
produced at E.
Proof.— AD || EC, /. L BAD = L AEC, (1—9, p. 42.)
and L DAC = L ACE. (1—8, p. 40.)
But L BAD = L DAC, by hypothesis,
/. L AEC = L ACE.
/. AC = AE. (1—13, p. 52.)
RATIO AND PROPORTION 231
In A EBC, AD || EC,
••§§ = fr (IV-2,p.222.)
But AE = AC,
J3D _ BA
'*• D~C ~ AC'
THEOREM 5
(Converse of Theorem 4)
If the base of a triangle is divided internally
into segments that are proportional to the other
sides of the triangle, the straight line which joins
the point of section to the vertex bisects the
vertical angle.
B D C
Hypothesis. — In A ABC, ^ = ^ •
To prove that AD bisects L BAC.
Construction. — Bisect L BAC and let the bisector cut
BC at E.
Proof. — v AE bisects L BAC
•"• f§ = f5' (IV— 4, p. 230.)
But, by hypothesis,
BE = BD
•"* EC ~ DC "
.*. E and D coincide.
.'. AD bisects L BAC.
232 THEORETICAL GEOMETRY BOOK IV
THEOREM 6
The bisector of the exterior vertical angle of a
triangle divides the base externally into segments
that are proportional to the sides of the triangle.
Hypothesis. — In A ABC, BA is produced to F.
L FAC is bisected by AD which cuts BC produced
at D.
BD BA
To prove ^D = AC"
Construction. — Through C draw CE II AD to meet
AB at E.
Proof.— ': EC || AD, .'. L FAD = Z AEC. (1—9, p. 42.)
and L DAC = L ACE. (1—8, p. 40.)
But, by hypothesis, Z FAD = L DAC.
/. L AEC = L ACE.
.-. AC = AE. (1—13, p. 52.)
In A BAD, EC II AD,
BA '= £D (IV— 2, Cor.,
" AE DC p. 223.)
But AE - AC.
BA = BD
•*• AC CD"
EXERCISES 233
THEOREM 7
(Converse of Theorem 6)
If the base of a triangle is divided externally so
that the segments of the base are proportional to
the other sides of the triangle, the straight line
which joins the point of section to the vertex
bisects the exterior vertical angle.
B C D
Hypothesis. — In A ABC, ^R = ^, and BA is produced
to E.
To prove that AD bisects L CAE.
Construction. — Bisect L EAC by AF.
Proof. — v AF bisects exterior L EAC,
... BF=|A (IV— 6, p. 232.)
\J r A\Q^
But, by hypothesis,
BF _ BD
*• CF ~ CD*
/. D and F coincide
.'. AD bisects L EAC.
125.— Exercises
1. The sides of a A are 4 cm., 5 cm., 6 cm. Calculate
the lengths of the segments of each side made by the
bisector of the opposite /_.
2. AD bisects L A of A ABC and meets BC at D. Find
BD and CD in terms of a, 6, and c.
234 THEORETICAL GEOMETRY BOOK IV
3. In A ABC, a = 7, b = 5, c = 3. The bisectors of the
exterior Z_s at A, B, C meet BC, CA, AB respectively at
D, E, F. Calculate BD, AE and AF.
4. In A ABC, the bisector of the exterior L at A meets
BC produced at D. Find BD and CD in terms of a, b and c.
5. If a st. line bisects both the vertical L and the base
of -a A, the A is isosceles.
6. The bisectors of the Z_s of a A are concurrent. (Use
IV_4 and 5.)
7. AD is a median of A ABC; L.S ADB, ADC are
bisected by DE, DF meeting AB, AC at E, F respectively.
Prove EF || BC.
8. The bisectors of Ls A, B, C in A ABC meet BC, CA,
AB at D, E, F respectively. Show that AF.BD.CE =
FB.DC.EA.
9. If the bisectors of L s A, C in the quadrilateral A BCD
meet in the diagonal BD, the bisectors of Z_s B, D meet in
the diagonal AC.
ylO. If the bisectors of Z_s ABC, ADC in the quadrilateral
ABCD meet at a point in AC, the bisectors of the exterior
Z_s at B and D meet in AC produced.
11. If O is the centre of the inscribed circle of A DEF
and DO produced meets EF at G, prove that DO : OG =
ED+ DF: EF.
12. PQ is a chord of a circle _L to a diameter MN
and D is any point in PQ. The st. lines MD, ND meet
the circle at E, F respectively. Prove that any two
adjacent sides of the quadrilateral PEQF are in the same
ratio as the other two.
13. The bisector of the vertical L of a A and the
bisectors of the exterior Z_s at the base are concurrent.
EXERCISES
235
14. One circle touches another internally at M. A chord
PQ of the outer circle touches the inner circle at T.
PT PM
PrOV6that ^-
15. LMN is a A in which LM = 3 LN. The bisector of
L L meets MN in D, and MX _L LD. Prove that
LD = DX.
16. The Z. A of A ABC is bisected by AD, which cuts
the base at D, and O is the middle point of BC. Show
that OD is to OB as the difference of AB and AC is to
their sum.
17. The bisectors of the interior and exterior L s at the
vertex of a A divide the base internally and externally in
the same ratio.
18. A point P moves so that the ratio of its distances
from two fixed points Q, R is constant. Prove that the
locus of P is £ circle. (The Circle of Apollonius.)
Divide QR internally at
S and externally at T so
that
QS _ QT _ PQ
SR ~ TR ~ PR'
Join PS, PT; and pro-
duce QP to V.
PQ
SR " PR'
£••_££ .-. z QPS = z SPR.
Z RPT = Z TPV.
QT _ PQ
TR T PR'
.-. Z SPT - QPS + Z TPV
= J st. Z QPV
= a rt. / "j
and, hence, a circle described on ST as diameter passes
through P.
19. If L, M, N be three points in a st. line, and P a
point at which LM and MN subtend equal L s, the locus
of P is a circle.
236 THEORETICAL GEOMETRY BOOK IV
SIMILAR TRIANGLES
THEOREM 8
If the angles of one triangle are respectively
equal to the angles of another, the corresponding
sides of the triangles are proportional.
vA
B F C E F
Hypothesis.— In As ABC, DEF; L A = ^ D, Z. B =
L E, L C = L F.
Proof. — Apply A DEF to A ABC so that L E coin-
cides with L B; the A DEF taking the position D'BF'.
.'. L BD'F' = L A, .'. D'F' || AC. (1—7, p. 38.)
/.£! = £! (IV-2, Cor., p. 223.)
AB _ BC
DE~EF
In the same way, by applying the As so that
Rf^ r"* A
C and F coincide, it may be proved that = p^
' DE~ EF~ FD
AB_BC AB DE
DE ~ EF' " BC = EF'
and in the same way ^ = ^ and 5^ = £P.
/. If two triangles are similar, the corresponding
sides about the equal angles are proportional.
• SIMILAR TRIANGLES 237
THEOREM 9
(Converse of Theorem 8)
If the sides of one triangle are proportional to the
sides of another, the triangles are similar, the
equal angles being opposite corresponding sides.
B
Hypothesis.— In As ABC, DEF; ^| = |^ = ?£.
To prove Z_ A = Z. D, L B = L DEF, L C = L DFE.
Construction. — Make L FEG = L B, L EFG = L C.
tL A = Z. G,
Proof.— In As ABC, GEF J A B = L GEF,
l^- C = L EFG.
.'. A ABC III A GEF.
.••ti = fr- (IV-8, p. 236.)
But, by hypothesis, = -
Similarly it may be proved that GF = DF.
./-DE = GE,
In As DEF, GEF J EF is common,
IFD - FG.
/. A DEF ~ A GEF. (1—4, p. 22.)
.'. L DEF = L GEF = L B, L DFE = L GFE = L C.
.% remaining L D = remaining L A.
238 THEORETICAL GEOMETRY BOOK IV
126.— Exercises
1. The st. line joining the middle points of the sides of
a A is || to the base, and equal to half of it.
2. If two sides of a quadrilateral be ||, the diagonals cut
each other proportionally.
3. In the A ABC the medians BE, CF cut at G. Show
that BG = twice GE, and CG = twice GF.
4. Using the theorem in Ex. 3, devise a method of
trisecting a st. line.
5. If three st. lines meet at a point, they intercept on
any || st. lines portions which are proportional to one
another.
6. In similar AS J_s from corresponding vertices to the
opposite sides are in the same ratio as the corresponding
sides.
7. In similar AS the bisectors of two corresponding Ls,
terminated by the opposite sides, are in the same ratio as
the corresponding sides.
Jtf&. A BCD is a ||gm, and a line through A cuts BD at
E, BC at F and meets DC produced at G. Show that
AE : EF = AG : AF.
j^9. If two || st. lines AB, CD be divided at E, F
respectively so that AE : EB = CF : FD, then AC, BD and
EF are concurrent.
10. The median drawn to a side of a A bisects all st.
lines || to that side and terminated by the other two sides,
or those sides produced.
^11. ABCD is a ||gm. AD is bisected at E and BC at F.
Show that AF and CE trisect the diagonal BD.
•*12. If the st. lines OAB, OCD, OEF be similarly divided,
the AS ACE, BDF are similar.
EXERCISES 239
13. If the corresponding sides of two similar As be ||,
the st. lines joining the corresponding vertices are con-
current.
14. A LMN HI A PQR, ^ L = L P and L M = L Q. LM =
7 cm., MN = 5 cm., LN = 9 cm., QR = 4 cm. Find PQ
and PR.
15. In A DBF, DE = 13 cm., EF = 5 cm. and DF - 12
cm. The A is folded so that the point D falls on the
point E. Find the length of the crease.
16. LMN is a A and X is any point in MN. Prove
that the radii of the circles circumscribing LMX, LNX are
proportional to LM, LN.
17. St. lines POQ, ROS are drawn so that PO = 2 OQ
and RO = 2 OS. RQ and PS are produced to meet at T.
Prove that PS ^ST and RQ = QT.
18. FDE, GDE are two circles and FDG is a st. line.
FE, GE are drawn. Prove that FE is to GE as diameter
of circle FDE is to diameter of GDE.
19. P is any point on either arm of an L XOY, and
P N
PN ± to the other arm. Show that - has the same
value for all positions of P.
Show also that Q-= has the same value for all positions
p N
of P ; and that Q-=^ has the same value for all positions
of P.
P N
(NOTE. — The ratio =-= is catted the sine of the L XOY,
Q-p is the COSine of that L, and ^-^ is the tangent of the
same ^-.
240 THEORETICAL GEOMETRY BOOK IV
20. PQRS is a quadrilateral inscribed in a -circle. The
diagonals PR, QS cut at X. Prove that ^9 = *?.
SR XS
21. OX, OY, OZ are three fixed st. lines, and P is any
point in OZ. From P, PL is drawn _L OX and PM J_
OY. Prove that the ratio PL : PM is constant.
22. In the quadrilateral DEFG the side DE || GF and
the diagonals DF, EG cut at H. Through H the line
LHM is drawn || DE and meeting EF, DG at L, M respec-
tively. Prove HL = HM.
23. KLMN is a quadrilateral in which KL || NM. Prove
that the line joining the middle points of KL and MN
passes through the intersection of the diagonals KM, LN.
24. DEF is a A and G is any point in EF. The
bisector of L DGF meets DF in H. EH cuts DG at
K. FK meets DE at L. Prove that LG bisects L DGE.
25. DG and DH bisect the interior and exterior L s at
D of a A DEF, and meet EF at G and H ; and O is the
middle point of- EF. Show that OE is a mean proportional
between OG and OH.
26. DG bisects L D of A DEF and meets EF at G.
GK bisects L DGE and meets DE at K. GH bisects
L DGF and meets DF at H. Prove that A EKH : A FKH
= ED : DF.
SIMILAR TRIANGLES 241
THEOREM 10
If two triangles have one angle of one equal to
one angle of the other and the sides about these
angles proportional, the triangles are similar, the
equal angles being opposite corresponding sides.
B C E F
Hypothesis. — In As ABC, DEF, L A = L D
To prove A ABC ||| A DEF.
Proof. — Apply the As so that L D coincides with
L. A and A DEF takes the position AE'F'.
.. AB AC
* DE ~ DF'
•••s! = ^ ••'E'F/»BC- <IV-3,p.224.)
/. LB= L AE'F, L C = L AF'E', (I— 9, p. 42.)
.'. A ABC. Ill A AE'F.
But A AE'F is the triangle DEF in its new position,
/. A ABC HI A DEF.
The equal ^.s B, E are respectively opposite the
corresponding sides AC, DF, also the equal Z_s C, F
are respectively opposite the corresponding sides AB,
DE.
242 THEORETICAL GEOMETRY BOOK IV
THEOREM 11
If two triangles have two sides of one propor-
tional to two sides of the other, and the angles
opposite one pair of corresponding sides in the
proportion equal, the angles opposite the other
pair of corresponding sides in the proportion are
either equal or supplementary.
Hypothesis.— IK As ABC, DEF, - jf and L B =
L E.
To prove either L C = L F or Z.C+Z. DFE = 2 rt. L. B.
Proof.— (I) If L A - L D. (Fig. 1.)
V L A = L D, and L B = L E, /. L C = L F.
In this case A ABC III A DEF.
(2) If L A is not equal to L D. (Fig. 2.)
At D make L EDG = L A and produce DG to meet
EF, produced if necessary, at G.
c L A = L EDG,
In As ABC, DEG J L B = L E,
I .'. L C = L G.
.'. A ABC HI A DEG.
-s- p- 236->
EXERCISES 243
But, by hypothesis, £| = 5^ •
...£ = £,„« -OF.
In A DFG, V DF = DG, .'. L DGF = Z DFG.
But Z DGF = Z C, /. L DFG = L C.
L DFE-f /- DFG = 2 rfc. ^ S,
/. L DFE + A C = 2 rt. z_ s.
127.— Exercises
1. Show that certain propositions of Book I are respec-
tively particular cases of Theorems 9, 10 and 11 of
Book IV.
2. In similar AS medians drawn from corresponding
vertices are proportional to the corresponding sides.
3. In a A ABC, AD is drawn J_ BC. If BD : DA =
DA : DC, prove that BAG is a rt. /_.
4. If the diagonals of a quadrilateral divide each other
proportionally, one pair of sides are ||.
5. A point D is taken within a A LMN and joined to
L and M. A A EMN is described on the other side of
MN from A LMN having L EMN * L DML, and L ENM =
L DLM. Prove that A DME 111 A LMN.
6. M, N are fixed points on the circumference of a given
circle, and P is any other point on the circumference.
MP is produced to Q so that PQ : PN is a fixed ratio.
Find the locus of Q.
7. EOD, GOF are two st. lines such that GO : DO =
EO : FO. Prove that E, F, D, G are concyclic.
8. OEF, OGD are two st. lines such that OE : OG =
OD : OF. Prove that E, F, G, D are concyclic.
9. DEF is a A, and FX J_ DE. Prove that, if DF:FX =
DE : EF, L XFE = L D.
244
THEORETICAL GEOMETRY
BOOK IV
10. Similar isosceles As DEF, DEG are described on
opposite sides of DE such that DF = DE and GD = GE.
H is any point in DF and K is taken in GD such that
GK : GD = DH : DF. Prove A KHE ||| A GDE.
11. LMN is a A, and D is any point in LM produced.
E is taken in NM such that NE : EM = LD : DM. Prove
that DE produced bisects LN.
12. O is the centre and OD a radius of a circle. E is
any point in OD, and F is taken' in OD produced such
that OF is a third proportional to OE, OD. P is any
point on the circumference. Prove L FPD = L DPE.
13. The bisectors of the interior and exterior z_s at L
in the A LMN meet MN and MN produced at D, E
respectively. FNG drawn || LM meets LE at F and LD
produced at G. Prove FN = NG.
14. If one pair of Z_s of two AS be equal and another
pair of Z_s be supplementary, the ratios of the sides
opposite to these pairs of Z_s are equal to each other.
GEOMETRIC MEANS
THEOREM 12
The perpendicular from the right angle to the
hypotenuse in a right-angled triangle divides the
triangle into two triangles which are similar to
each other and to the original triangle.
B D C
Hypothesis. — In A ABC, L BAG is a rt.
AD ± BC.
L and
GEOMETRIC MEANS 245
To prove A ABD ||| A CAD III A CBA.
Proof. — f Z B is common.
In As ABD, CBA \ L BDA = L BAC, both rt. L s.
[ /. L BAD = Z. BCA.
/. A ABD III A CBA
Similarly A ADC ||| A CBA.
/. AABD III A CAD HI A CBA.
Cor. 1.— V A ABD III A QAD,
BD AD
AD ~CD'
.*. AD is the mean proportional between BD and DC.
Cor. 2. — Because A ABD ||| A CBA
BD AB
AB ~ BC*
.'. AB is the mean proportional between BD and BC.
Similarly —
AC is the mean proportional between DC and CB.
Cor. 3. — Because A CBA 111 A CAD,
CB = CA
BA ~ AD"
i.e., the hypotenuse is to one side as the other side
is to the perpendicular.
246 THEORETICAL GEOMETRY BOOK IV
PROBLEM 5
To find the mean proportional between two given
straight lines.
From a st. line cut off AB, BC respectively equal to
the two given st. lines.
It is required to find the mean proportional to AB, BC.
On AC as diameter describe a semi-circle ADC.
From B draw BD _|_ AC and meeting the arc ADC
at D.
BD is the required mean proportional.
Join AD, DC.
Probf. — v ADC is a semi-circle,
.'. L ADC is a rt. L. (Ill— 9, p. 160.)
In A ADC, L ADC is a rt. L,
and DB _L AC.
•••^ = §|- (IV-12, Cor. 1, p. 245.)
.'. BD is the mean proportional between
AB and BC.
• RECTANGLES 247
RECTANGLES
THEOREM 13
If four straight lines are proportionals, the
rectangle contained by the means is equal to the
rectangle contained by the extremes.
A
B.
c.
D.
B A
Hypothesis. — A, B, C, D are four st. lines such that
A C
B ~ D"
To prove that rect. B.C = rect. A.D.
Let a, b, c, d be the numerical measures of A, B,
C, D respectively.
Then
/. be — ad.
But be is the numerical measure of B.C and ad is
the numerical measure of A.D,
.'. rect. B.C = rect. A.D.
248
THEORETICAL GEOMETRY
THEOREM 14
BOOK IV
(Converse of Theorem 13)
If two rectangles are equal to each other, the
length of one is to the length of the other as the
breadth of the second is to the breadth of the first.
B c F G
Hypothesis. — Reel ABCD = rect. EFGH.
To prove
BC
FG
EF
AB
Proof. — Let a, b, c, d be the numerical measures of
BC, BA, FG, EF respectively.
Then since the rectangles are equal,
ab - cd.
d
b'
BC
FG
EF
AB
THE PYTHAGOREAN THEOREM
249
Alternative proof of the Pythagorean Theorem.
—is, p. m.)
The square on the hypotenuse of a right-angled
triangle equals the sum of the squares on the other
two sides.
Hypothesis. — BAG is a A having L BAG a rt. Z,
and having squares described on the three sides.
To prove that BC2 = BA2 + AC2.
Construction. — Draw AD _|_ BC.
Proof. — v BAC is a rt.-^-d A with AD _L the hypo-
tenuse BC,
if = §5' (IV-12,Cor.2,p.245.)
BA2 = BC.BD. (IV— 13, p. 247.)
Similarly CA2 - BC.CD.
.'. BA2 4- CA2 = BC.BD 4- BC.CD
= BC (BD 4- CD)
= BC.BC
= BC*
i.e., BC2 = BA2 4- CA2.
250
THEORETICAL GEOMETRY
BOOK IV
128.— Exercises
1. Give a general enunciation of IY — 12, Cor. 1.
2. Give a general enunciation of IY — 12, Cor. 2.
3. Give an alternative proof of IY — 13, using the con-
struction indicated in the following diagram : —
'N M
A B
-D
AB EF
QQ = — • In the rectangles NL, RL, KL = AB, LM = GH,
PL = CD and LQ - EF.
Using a similar construction give also an alternative proof
of IY— 14.
4. In any two equal As ABC, DEF, if AG, DH be J_s
to BC, EF respectively, AG : DH = EF : BC.
5. In any A the J_s from the vertices to the opposite
sides are inversely as the sides.
6. In the diagram of IY — 12, show that rect. AD.BC =
rect. BA.AC. Give a general statement of this theorem.
7. ABC, DEF are two equal As having also LB = LE.
Show that f^ = £|-
8. ABCD, EFGH are two equal ||gms having also LB =
Z.F. Show that 5*? = £| .
9. ABCD is a given rect. and EF a given st. line. It is
required to make a rect. equal in area to ABCD and
having one of its sides equal to EF.
EXERCISES 251
10. Make a rect. equal in area to a given A and having
one of its sides equal to a given st. line.
11. Show how to construct a rect. equal in area to a
given polygon and having one of its sides equal to a given
st. line.
12. If from any point on the circumference of a circle a
J_ be drawn to a diameter, the square on the _L equals
the rect. contained by the segments of the diameter.
13. Construct a square equal to a given rect.
14. Construct a square equal to a given ||gm.
15. Construct a square equal to a given A.
16. Draw a square having its area 12 sq. inches.
17. Divide a given st. line into two parts such that the
rect. contained by the parts is equal to the square on
another given st. line.
18. If a st. line be divided into two parts, the rect.
contained by the parts is greatest when the line is bisected.
19. AB and C are two given st. lines. Find a point D
in AB produced such that rect. AD.DB = sq. on C.
20. Construct a rect. equal in area to a given square
and having its perimeter equal to a given st. line.
When will the solution be impossible?
21. Show how to construct a square equal in area to a
given polygon.
22. In the corresponding sides BC, EF of the similar AS
ABC, DEF the points G, H are taken such that BG : GC =
EH : HF. Prove AG : DH = BC : EF.
252 THEORETICAL GEOMETRY BOOK IV
CHORDS AND TANGENTS *
THEOREM 15
If two chords intersect within a circle, the
rectangle contained by the segments of one is
equal to the rectangle contained by the segments
of the other.
B
Hypothesis. — In the circle ABC, the chords AC, BD
intersect at E.
To prove that rect. AE.EC = rect. BE. ED.
Construction. — Join AB, CD.
Proof. — v Z_s ABD, ACD are in the same segment,
.'. L ABD = L ACD. (Ill — 7, p. 156.)
Similarly, L BAC = L BDC.
And L AEB = L CED. (I — 1, p. 13.)
/. A AEB HI A DCE.
•'• ED = §5" (IV-8, p. 236.)
.'. rect. AE.EC = rect. BE. ED. (IV— 13, p. 247.)
CHORDS AND TANGENTS 253
THEOREM 16
(Converse of IV— 15)
If two straight lines cut each other so that the
rectangle contained by the segments of one is equal
to the rectangle contained by the segments of the
other, the four extremities of the two straight lines
are concyclic.
Hypothesis — The st. lines AB, CD cut at E so that
rect. AE.EB = rect. CE.ED.
To prove that A, C, B, D are concyclic.
Construction. — Describe a circle through A, C, B, and
let it cut ED, produced if necessary, at F.
Proof. — v AB, CF are chords of a circle,
.'. AE.EB = CE.EF. (IV— 15, p. 252.)
But, AE.EB = CE.ED. (Hyp.)
:. CE.EF = CE.ED.
And /. EF = ED.
/. F coincides with D,
and the points A, C, B, D are concyclic.
254 THEORETICAL GEOMETRY BOOK IV
•t \
THEOREM 17 "<
If from a point without a circle, a secant and a
tangent are drawn, the square on the tangent is
equal to the rectangle contained by the secant, and
the part of it without the circle.
Hypothesis. — PA is a tangent and PCB a secant to
the circle ABC.
To prove that PA2 = PB.PC.
Construction. — Join AB, AC.
Proof. — V AP is a tangent, and AC is a chord from
the same point A,
.V L PAC = L ABC. (Ill— 15, p. 177.)
L P is common,
In As PAB, PCA,\ L PBA = L PAC,
and .*. , L PAB = L PCA,
/. A PAB III A PCA.
••-£! = ££• (IV-8, p. 236.)
/. PA2 = PB.PC. (IV— 13, p. 247.)
CHORDS AND TANGENTS 255
THEOREM 18
(Converse of IV— 17)
If from a point without a circle two straight lines
are drawn, one of which is a secant and the other
meets the circle so that the square on the line
which meets the circle is equal to the rectangle
contained by the secant and the part of it without
the circle, the line which meets the circle is a
tangent.
"A
Hypothesis. — PA and PBC are drawn to the circle
ABC so that PA2 = PB.PC.
To prove that PA is a tangent.
Construction. — Join AB, AC.
Proof. — In As PAB, PAC, Z P is common,
and v PA2 = PB.PC,
^ = I** ' (IV— 14, p. 248.)
.'. A PAB in APCA. (IV— 10, p. 241.)
/. Z PAB = Z PCA.
/. PA coincides with the tangent
at A. (Ill— 15, p. 177.)
i.e., PA is a tangent to the circle.
NOTE. — Prove this proposition with the folloiving construc-
tion : — Draw a tangent from P, and join the point of contact
and the points A, P to the centre.
256 THEORETICAL GEOMETRY BOOK IV
129.— Exercises ,
1. PAB, PCD are two secants drawn from a point P
without a circle. Show that rect. PA.PB = rect. PC.PD.
From this exercise deduce a proof for IV — 17.
2. If in two st. lines PB, PD points A, C respectively be
taken such that rect. PA.PB = rect. PC.PD, the four
points A, B, C, D are concyclic.
3. If two circles intersect, their common chord bisects
their common tangents.
^ 4. If two circles intersect, the tangents drawn to them
from any point in their common chord produced are equal
to each other.
-^jf- 5. Through P any point in the common chord, or the
common chord produced, of two intersecting circles two
lines are drawn cutting one circle at A, B and the other
at C, D. Show that A, B, C, D are concyclic.
/$. Through a point P within a circle, any chord APB is
drawn. If O be the centre, show that rect. AP.PB =
OA2 - OP2.
/ 7. From any point P without a circle any secant PAB
is drawn. If O be the centre, show that rect. PA.PB =
OP2 - OA2.
8. From a given point as centre describe a circle cutting
a given st. line in two points, so that the rectangle con-
tained by their distances from a given point in the st.
line may be equal to a given square.
• 9. Describe a circle to pass through two given points and
touch a given st. line.
10. If three circles be drawn so that each intersects the
other two, the common chords of each pair meet at a point.
EXERCISES 257
11. Find a point D, in the side BC of A ABC, such
that the sq. on AD = rect. BD.DC. When is the solution
possible ?
12. Use IV — 17 to find a mean proportional to two
given st. lines.
13. P is a point at a distance of 7 cm. from the centre
of a circle. PDE is a secant such that PD = 5 cm. and
DE = 3 cm. Find the length of the radius of the circle.
14. In a circle of radius 4 cm. a chord DE is drawn 7
cm. in length. F is a point in DE such that DF = 5 cm.
Find the distance of F from the centre of the circle.
15. DEF is an isosceles A in which ED = EF. A circle,
which passes through D and touches EF at its middle
point cuts DE at H. Prove that DH = SHE.
16. In a circle two chords DE, FG cut at H. Prove that
(FH - HG)2 — (DH - HE)2 = FG2— DE2.
17. LND, MNE are two chords intersecting inside a
circle and LM is a diameter. Prove that
LN.LD-f MN.ME = LM2.
18. DEF, HGF are two circles and DFG is a fixed st.
line. Show how to draw a st. line EFH such that EF.FH
= DF.FG.
19. P is a point in the diameter DE of a circle, and PT
is the J_ on the tangent at a point Q. Prove that
PT.DE = DP.PE+ PQ2-
20. P, Q, R, S are four points in order in the same st.
line. Find a point O in this st. line such that OP. OR =
OQ.OS.
21. The tangent at P to a circle, whose centre is O meets
two || tangents in Q, R. Prove that PQ.PR = OP2.
258 THEORETICAL GEOMETRY BOOK IV
Miscellaneous Exercises -,
1. EFGH is a ||gm, P a point in EF such that EP:PF =
mm. What fraction is A EPH of the ||gm?
2. EFGH is a ||gm, P is a point in the diagonal FH
such that FP:PH -2:5. What fraction of the ||gm is A
EFP? If FP:PH = m:n find the fraction.
3. EFGH is a ||gm, P is a point in the diagonal FH
produced such that FP : PH = 9:5. What fraction of the
||gm is the A PEH ?
4. KLMN is a ||gm. Any st. line EKG is drawn cutting
the sides ML and MN produced at E and G. Show that
half the ||gm is a mean proportional between As EKL and
NKG.
5. The A PQR has PQ and QR divided at D and E
such that PD : DQ = QE : ER = 1 : 3. PE and RD intersect
at O. Find the ratios of the As PDO : OPR :OER : PQR.
6. D arid E are points in PQ and PR sides of the A
PQR such that QD : DP = PE : ER = m:n. Compare the
areas of the As QDE and DER.
7. Either of the complements of the ||gms about the
diagonal of a ||gm is a mean proportional between the two
||gms about the diagonal.
8. LMN is an isosceles A having LM = LN, LD is per-
pendicular to MN, P is a point in LN such that LP : LM
= 1:3. Prove that MP bisects LD.
9. Through E one of the vertices of a rectangle EFGH
any st. line is drawn, and HP and FQ are J_s to PEQ.
Prove PE.EQ = HP.FQ.
10. DEF is a A, P and Q are points in DE and DF,
and DP : PE = 3 : 5 and DQ : QF = 7:8. In what ratio is PQ
cut by the median DG ?
MISCELLANEOUS EXERCISES 259
11. DEFG is a ||gm, and EF is produced to K so that
FK = EF; DK cuts EG at P. Show that GP = & EG.
12. The diagonals of the ||gm EFGH intersect at O ; if
E be joined to the middle point P of OH, and EP and FG
meet at K, find GK:EH.
13. DEF is a right-angled A, E being the right angle.
G is taken in DE produced such that DG:GF = DF : EF.
Prove that Z DFG is right.
14. If the perpendicular to the base of a A from the
vertex be a mean proportional to the segments of the base,
the triangle is right angled.
15. DGH is any A, and from K the middle point of GH
a line is drawn cutting DH at E and GD produced at F.
Prove GF:FD = HE: ED. Prove the converse also.
16. AD and AE are the interior and exterior bisectors of
the vertical angle of A ABC meeting the base at D and
E. Through C, FCG is drawn || to AB meeting AD and
AE at F and G. Prove that FC = CG.
17. HKL is »n isosceles A, having HK = HL; KL is
produced to D and DEF is drawn cutting HL at E, and
HK at F. Prove DE:DF = EL:KF.
18. DP and DQ are perpendiculars to the bisectors of
the interior angles E and F of any A DEF. Prove PQ || EF.
10. PX and QY are perpendiculars from P and Q to
XY; PY and QX intersect at R, and RZ is perpendicular
to XY. Prove Z PZX = Z QZY.
20. ABC is any A, and AD is taken along AC such
that AC : AB = AB : AD ; also CF is taken along AC such
that AC:CB = CB:CF. Prove BF = BD.
21. The perpendicular KD to the hypotenuse HL of a
right-angled A KHL is produced to E such that KD : DH
= DH:DE. Prove HE || KL.
260 THEORETICAL GEOMETRY BOOK IV
22. DEF is a A inscribed in a circle, and. P and Q are
taken in DE and DF such that DP : PE = DQ QF. Show
that the circle described about D, P, Q touches the given
circle at D.
23. D is a point in LM a side of A LMN, DE is || to
MN and EF || to LM, meeting the sides at E and F.
Prove LD:DM = MF:FN.
24. A variable line through a fixed point O meets two
|| st. lines at P and Q. Prove OP:OQ a constant ratio.
25. If the nonparallel sides of a trapezium are cut in
the same ratio by a st. line, show that this line is || to
the || sides.
26. ABODE is a polygon, O a point .within it. If X, Y,
Z, P, Q are points in OA, OB, OC, OD, OE such that OX:
OA - OY:OB = etc., show that the sides of XYZPQ are ||
to those of ABODE.
27. DE is a st. line, F any point in it ; find a point P
in DE produced such that PD:PE = DF:FE.
28. St. lines PD, PE, PF and PG are such that each
of the z_s DPE, EPF, FPG is equal to half a right angle.
DEFG cuts them such that PD = PG. Prove that DG : FG
- FG : EF.
29. GH is a chord of a circle, K and D points on the
two arcs respectively ; KH and KD are joined and GD
meets KH produced at E; EF || to GH meets KD produced.
Show that EF is equal to the tangent from F.
30. DEF, DEG are two circles, the centre P of DEG
being on the circumference of DEF. A st. line PHGF cuts
the common chord at H. Prove that PH : PG — PG : PF.
31. EF is the diameter of a circle. PQ is a chord J_
to EF, a chord QXR cuts EF at X, and PR, EF produced
meet at Y. Show that EX : EY = FX : FY.
MISCELLANEOUS EXERCISES 261
32. O is a fixed point and P a variable point on the
circumference of a circle; PO is produced to Q such that
OQ:OP = ra 7i. Find the locus of Q.
33. LMN is a A inscribed in a circle, L L is bisected
by LED cutting MN at E and the arc at D. Prove As
LEN and LMD similar.
34. The L D of the A DEF is bisected by DP cutting
EF in P; QPR is JL to DP meeting DE and DF at Q
and RS RS is || to EF meeting DE at S. Prove SE = EQ.
35. AOB, COD and EOF are any three st. lines; ACE
is || to FOB. Prove AC:CE = BD:DF. State and prove
a converse to this theorem.
36. Two circles DEF and DEG intersect; a tangent DF
is drawn to DEG, and EG to DEF. Show that DE is a
mean proportional between FE and DG.
37. EFGH is a quadrilateral, the diagonals EG and FH
meet at Q. Prove A EFH : A FGH = EQ : QG.
38. EFGH is a quadrilateral of which the sides EH and
FG produced meet at P. Prove A EFG : A FGH = EP :
PH.
39. G is the middle point of the st. line MN, PE a st.
line || to MN. Any st. line EFGH cuts PN at F and PM
produced at H. Prove EF : FG = EH : HG.
40. ABC is a A having Z B = Z C = twice Z A, BD
bisects the Z B meeting AC at D. Prove AC : AD = AD:
DC; also prove A ABC : A ABD = A ABD:A BDC.
41. EFGH is a cyclic quadrilateral, EG and FH intersect
at O, and OP and OQ are _|_s to EH and FG. Show
that OP:OQ = EH : FG.
42. EF is the diameter of a circle and P and Q any
points on the circumference on opposite sides of EF; QR is
J_ to EF meeting EP at S. Prove A ESQ ||| A EQP.
262 THEORETICAL GEOMETRY BOOK IV
43. ABC is a A inscribed in a circle, centre O, AD a
J_ to BC, AOE a diameter. Prove As ADC and ABE
similar: and AD.AE = AB.AC.
44. EFG is a A inscribed in a circle, ED || to the tangent
at G meets the base at D. Prove that FG : FE = EG : ED.
45. Find the ratio of the segments of the hypotenuse of a
right- 2_d A made by a perpendicular on it from the vertex,
if the ratio of the sides be (1) 1 : 2 ; (2) m : n.
46. PQ is the diameter of a circle; a tangent is drawn
from a point R on the circumference, PS and QT are J_ to
the tangent. Prove As PRQ, RPS and RTQ similar; also
show that A PRQ is half of PSTQ.
47. PQ and PR are tangents to a circle, PST is a secant
meeting the circle at S and T. Prove QT : QS = RT : RS.
48. Two circles intersect at E and F ; from P, any point on
one of them, chords PED, PFG are drawn, E£ and DG meet
at Q and PQ cuts the circle PEF at R. Prove R, F, G, Q
concyclic ; also that PQ2 is equal to the sum of the squares on
the tangents to the circle EFGD from P and Q.
49. PBR is a st. line, and similar segments of circles, PAB
and BAR, are described on PB and BR and on the same
side of PR. PAC and RAD are drawn to meet the circles
at C and D. Prove PD : RC = PB : BR.
NOTE. — Segments of circles are said to be similar when they
contain equal angles.
50. PMQ is the diameter of a circle PRQ, PX and QY
are || tangents, XRY is any other tangent, PY and XQ
meet at O. Show that RO is j| to PX ; that RO produced
to M is J_ to the diameter; and that MO = OR.
51. ABCD is a rectangle, a st. line APQR is drawn
cutting BC at P, the circle circumscribing the rectangle
at Q and DC produced at R, and such that AC bisects
L DAR. Prove DC : CR = PQ : PA.
MISCELLANEOUS EXERCISES 263
52. PQRS is a square. A st. line PFED cuts QS at F,
SR at E and QR produced at D. Prove FR a tangent to
the circle described about DER ; also that EF : PF = PF : FD.
53. FGHK is a cyclic quadrilateral, the L GFE is made
equal to L HFK and E is in GK. Prove As FEK and
FGH similar.
54. PA and PB are tangents to a circle, centre O, AB
meets PO in R ; PCD is any secant, OS is J_ to PD, and
AB and OS produced meet at Q. Prove (1) P, R, S, Q
concyclic; (2) PO.OR = OA2; (3) QD and QC are tangents
to the given circle.
55. DEF is a A and P and Q are points in ED and FE
such that EP : PD = FQ : QE, and PQ meets DF produced at
R. Prove RF : RD = PE2 : PD2. (Through F draw a st. line \\
to DE to meet PR.)
56. If a square is inscribed in a rt.-z_d A having one side
on the hypotenuse, show that the three segments of the base
are in continued proportion.
57. FGH is a A and L G and L H are bisected by st. lines
which cut the opposite sides at D and E; if DE is || to GH,
then FG = FH.
58. From P, the middle point of an arc of a circle cut off
by a chord QR, any chord PDE is drawn cutting QR at D.
Show that PQ2 = PD.PE.
59. Draw a st. line through a given point so that the
perpendiculars on it from two other given points may be
(1) equal, (2) one twice the other, (3) three times the
other, (4) in a given ratio.
60. LMN is an isosceles A, the base MN is produced
both ways, in NM produced any point P is taken, and
in MN produced NQ is taken a third proportional to PM
and LM. Prove As PLQ and PLM similar.
264 THEORETICAL GEOMETRY BOOK IV
61. EDOF is the diameter of a circle, centre O. PE and
PG are tangents to the circle; GD is J_ to' EF. Prove
GD:DE = OE:EP.
62. DEF is a A inscribed in a circle, centre O. The
diameter J_ to EF cuts DE at P and FD produced at Q.
Prove As EPO and FOQ similar; and hence OE2 = OP.OQ.
63. ABC is a A inscribed in a circle. The exterior J_ at
A is bisected by a st. line cutting BC produced at D and
the circumference at E. Prove BA.AC = EA.AD.
64. EFGH is a cyclic quadrilateral, P a point on the cir-
cumference, PQ, PR, PS, PT are J_ to EF, FG, GH, HE re-
spectively. Prove As PTQ and PSR similar; and PT.PR =
PS.PQ.
65. Any three || chords AB, CD, EF are drawn in a circle,
AC and BD meet EF produced at Q and R, P is a point in
the arc EF, and PA and PD meet EF at M and N. Prove
As AQM and NDR similar; hence show that, for all positions
of P, QM.NR is constant.
66. Two tangents TMP and TNQ are drawn to a circle,
centre O, and the st. line POQ is J_ to TO. MN is any other
tangent to the circle. Prove As MPO and NQO similar.
67. DH is a median of the A DEF, PQ is || to EF cutting
DE at P and DF at Q. Show that PF and EQ intersect
on DH.
68. LNM is a A inscribed in a semicircle, diameter LM.
NM is greater than NL. On opposite sides of LN the Z LNP is
made equal to /LNQ, P and Q lying along LM. Prove
PL : LQ = PM : QM.
69. EFGH is a ||gm, and RS is drawn || to HF meeting EH
and EF at R and S. Show that RG and SG cut off equal
segments of the diagonal FH. Prove a converse of this.
MISCELLANEOUS EXERCISES 265
70. ABC is a A and AB, AC are produced to D, E so that
BD = CE; DE and BC produced meet at F. Show that
AD : AE = FC : FB.
71. Two circles, centres O, P intersect, the centre O
being on the circumference of the other circle. GDE
touches the circle with centre O at G and cuts the other
at D, E, and EPF is a diameter. Prove A OGD III A
OEF; and hence, that OD.OE is constant for all positions
of the tangent.
72. Two circles touch externally at P; EF a chord of
one circle touches the other at D. Prove PE : PF = ED:DF.
73. EOF is the diameter of a circle, with centre O,
DP anj7- chord cutting the diameter; OSQR _L to DP meets
DP at S, DE at Q, and PE at R. Prove As EOF and
RSP similar; also OQ.OR = CD2.
74. Divide an arc of a circle into two parts so that the
chords which cut them off shall have a given ratio to each
other.
75. LMN is a A, and XY || MN meets LM at X and LN at
Y; MN is produced to D so that ND = XY, and XP || to LD
meets MN at P. Prove MN : ND = ND : NP.
76. Two circles intersect and a st. line CDOEF cuts the
circumferences at C, D, E, F and the common chord at O.
Show that CD : DO = EF : OE
77. DX _L EF and EY _[_ DF in A DEF. The lines DX,
EY cut at O. Prove that EX : XO = DX : XF.
78. From a point P without a circle two secants PKL,
PMN are drawn to meet the circle in K, L, M, N. The bi-
sector of L KPM meets the chord KM at E and the chord
LN at F. Prove that LF : FN = ME : EK
79. QR is a chord |j to the tangent at P to a circle. A
chord PD cuts QR at E. Prove that PQ is a mean propor-
tional between PE and PD.
THEORETICAL GEOMETRY BOOK IV
80. DEF, DEG are two fixed circles and PEG is a st. line.
Show that the ratio FD : DG is constant for all positions of
the st. line PEG.
81. DEF is a st. line, and EG, FH are any two || st. lines
on the same side of DEF such that EG : FH = DE : DF. Prove
that D, G, H are in a st. line.
82. From a given point on the circumference of a circle
draw two chords which are in a given ratio and contain a
given L.
83. DEF is a A and on DE, DF two As OLE, DFM are
described externally such that L FDM = L EDL and L DFM =
L OLE. Prove A DLF ||| A DEM.
84. DEFG is a ||gm and P is any point in the diagonal
EG. The st. line KPL meets DE at K and FG at L, and
MPN meets EF at M and GD at N. Prove KM || NL.
85. ABCD is a ||gm and PQ is a st. line || AB. The st.
lines PA, QB meet at R and PD, QC meet at S. Prove
RS || AD.
86. If the three sides of one A are respectively _L to
the three sides of another A> the two As are similar.
87. Find a point whose ± distances from the three
sides of a A are in the ratio 1 ; 2 ; 3.
88. Squares are described each with one side on one
given st. line and one vertex on another given st. line.
Find the locus of the vertices which are on neither.
89. If the sides of a rt.-/-d A are in the ratio 3 : 2,
prove that the JL from the vertex of the rt. L to the
hypotenuse divides it in the ratio 9 \ 4.
90. HK is a diameter of a circle and L is any point on
the circumference. A st. line J_ HK meets HK at D, HL
at E, KL at G, and the circumference at F. Show that
DF2 = DE.DG.
MISCELLANEOUS EXERCISES 267
91. The st. line joining a fixed point to any point on
the circumference of a given circle is divided in a given
ratio at P. Prove that the locus of P is a circle.
92. DEFG is a quadrilateral and P, Q, R, S are points on
DE, EF, FG, GD such that DP : DE = FQ : FE = FR : FG =
DS : DG. Prove that PQRS is a ||gm.
93. DEFG is a ||gm, and a line is drawn from E cutting
DF in P, DG in Q and FG produced in R. Prove that
PQ: PR = DP2: PF'; and that PQ.PR = EP2.
94. If A DEF:AGHK = DE.EFrGH.HK, prove that Zs
E, H are either equal or supplementary.
9">. From a point P without a circle draw a secant PQR,
such that QR is a mean proportional between PQ and PR.
96. Through a point of intersection of two circles draw a
line such that the chords intercepted by the circles are in a
given ratio.
97. If two As are on equal bases and between the same
|| s, the intercepts made by the sides of the As on any st.
line || to the base are equal.
98. The radius of a fixed circle is 38 mm., and a chord LM
of the circle is divided at P such that LP.PM = 225 sq. mm.
Construct the locus of P.
99. If the tangents from a given point to any number of
intersecting circles are all equal, all the common chords of
the circles pass through that point.
100. Circles are described passing through two fixed points ;
find the locus of a point from which the tangents to all the
circles are equal.
101. DEF is a A having Z E a rt. Z- A circle is de-
scribed with centre D and radius DE ; from F a secant is
drawn cutting the circle at G, H; and EX is drawn J_ DF.
Show that D, X, G, H are concyclic.
268 THEORETICAL GEOMETRY BOOK IV
102. GD is a chord drawn || to the diameter LM of a circle.
LG, LD cut the tangent at M at E, F respectively. Prove
that LG.GE-f LD.DF - LM2.
103. LM is a diameter of a circle, and on the tangent at
L equal distances LP, PQ are cut off. MP, MQ cut the
circumference at R, S respectively. Prove that LR : RS =
LM : MS.
104. GH drawn in the A DEF meets DE in G and DF
in H. From D any line DLK is drawn cutting GH in
L and EF in K. From L the st. lines LM, LN are
drawn || KH, KG and meeting DH, DG at M, N re-
spectively. Prove A LMN ||| A KHG.
105. In a given A inscribe an equilateral A so as to
have one side || to a side of the given A.
106. In a given A DEF draw a st. line PQ || ED
meeting EF in P arid DF in Q, so that PQ is a mean
proportional between EP and PF.
107. Two circles intersect at E, F, and DEG is the st.
line J_ EF and terminated in the circumferences. HEK
is any other st. line through E terminated in the circum-
ferences. HF, DF, KF, GF are drawn. Prove, by similar
As, that DG > HK.
108. In A ABC the bisectors of Z A and of the
exterior Z at A meet the st. line BC at D and E. Show
109. If two circles intercept equal chords PQ, RS on any
st. line, the tangents PT, RT to the circles at P, R are to one
another as the diameters of the circles.
110. DEF is a A having DF > DE. From DF a part
DG is cut off equal to DE, and GH is drawn || DE to meet
MISCELLANEOUS EXERCISES 269
EF at H. From GF a part GK is cut off equal to GH, and
KL is drawn || GH to meet EF at L; etc. Prove that DE,
GH, KL, etc., are in continued proportion.
1 1 1 . A circle P touches a circle Q internally, and also
touches two |j chords of Q. Prove that the J_ from the
centre of P on the diameter of Q which bisects the chords is
a mean proportional between the two extremes of the three
segments into which the diameter is divided by the chords.
1 1 2. PX is the J_ from a point P on. the circumference of
a circle to a chord QR, and QY, RZ are J_s to the tangent
at P. Prove that PX2 = QY.RZ.
113. Prove, by using 112, that if J_s are drawn to the sides
and diagonals of a cyclic quadrilateral from a point on the
circumference of the circumscribed circle, the rectangle con-
tained by the J_s on the diagonals is equal to the rectangle
contained by the J_s on either pair of opposite sides.
114. The projections of two || st. lines on a given st. line
are proportional to the st. lines.
115. DEFG is a square, and P is a point in GF such tnat
DP - FP-f- FE. Prove that the st. line from D to the middle
point of EF bisects L PDE,
116. DEF, GEF are As on opposite sides of EF, and DG
cuts EF at H. Prove that A DEF : A GEF = DH : HG.
117. From the intersection of the diagonals of a cyclic
quadrilateral J_s are drawn to a pair of opposite sides : prove
that these J_s are in the same ratio as the sides to which
they are drawn.
118. P, Q, R, S are points in a st. line, PX || QY, RX || SY,
and XY meets PS at O. Prove that OP.OS = OQ.OR.
119. From a point T without a circle tangents TP, TQ
and a secant TRS are drawn. Prove that in the quadri-
lateral PRQS the rect. PR.QS = the rect. RQ.SP.
BOOK V
AREAS OF SIMILAR FIGURES
THEOREM 1
The areas of similar triangles are proportional to
the squares on corresponding sides.
C EK F
Hypothesis. — ABC, DEF are similar As of which
BC, EF are corresponding sides.
To prove that
Construction.— Draw AH _L BC and DK _L EF.
Proof.— V A AHC III A DKF,
And A ABC ||| A DEF,
AH AC BC ,jy _ o n
~~
DK~DF~EF
A ABC = 4 AH.BC, (II — 4, p. 100.)
A DEF = 4 DK.EF,
A ABC 4 AH.BC
A~DEF 4 DK.EF
AH BC
~ DK'EF
BC BC^
~ EF'EF
BC2
~ EF"2"
271
272 THEORETICAL GEOMETRY BOOK V
130.— Exercises
«
1. Two similar As have corresponding sides in the ratio
of 3 to 5. What is the ratio of their areas'?
2. The ratio of the areas of two similar As equals the
ratio of 64 to 169. What is the ratio of their corresponding
sides 1
3. Draw a A having sides 4 cm , 5 cm., 6 cm. Make a
second A having its area four times that of the first, and
divide it into parts each equal and similar to the first A.
4. Show that the areas of similar As are as : —
(a) the squares on corresponding altitudes;
(b) the squares on corresponding medians ;
(c) the squares on the bisectors of corresponding /.s.
5. ABC, DEF are two similar As such that area of A
DEF is twice that of A ABC. What is the ratio of
corresponding sides'?
Draw A ABC having sides 5 cm., 6 cm., 7 cm., and
make A DEF similar to A ABC, and of double the area.
/ §. If ABC, DEF be similar As of which BC, EF are
corresponding sides, and the st. line G be such that
BC : EF = EF : G, then A ABC : A DEF = BC : G ; that is :—
If three st. lines be in continued proportion, the first is
to the third as any A on the first is to the similar A
similarly described on the second.
NOTE. — Similar As are said to be similarly described on
corresponding sides.
7. ABC is a A and G is a st. line. Describe a A DEF
similar to A ABC and such that A ABC : A DEF — BC : G.
Describe another A HKL similar to A ABC and such that
A ABC : A HKL = AB : G.
EXERCISES 273
8. Bisect a given A by a st. line drawn || to one of its
sides.
9. From a given A cut off a part equal to one-third of
its area by a st. line drawn || to one of its sides.
10. Trisect a given A by st. lines drawn || to one of its
sides.
11. Show that the equilateral A described on the hypote-
nuse of a rt.-/_d A equals the sum of the equilateral As
on the two sides.
12. In A DEF, DX JL EF and EY J_ FD. Prove that A
FXY : A DEF = FX2 : FD2.
13. In the acute-Ld A DEF, DX _L EF, EY _L FD,
FZ J_ DE, YG J_ EF and ZH J_ EF. Prove that XY and
XZ divide the A DEF into three parts that are pro-
portional to FG, GH and HE.
14. LMN is an equilateral A- The st. lines RLQ, PMR,
QNP are respectively _L LM, MN, NL. Find the ratio of
A PQR to A LMN.
15. A point O is taken in the diameter PQ produced of
a circle. OT is a tangent, and the tangent at P cuts OT
at N. If D is the centre of the circle, prove that
A OPN : A OTD = OP : OQ.
1 6. H is a point on the circumference of a circle of
which FG is a diameter, and O is the centre. HD _L FG,
and tangents at F and H meet at E. Prove that A FEH :
AOHG = FD : DG.
17. DEF, LMN are two As in which L E = L M. Prove
that A DEF : A LMN = DE.EF : LM.MN.
18. Similar As are to one another as the squares on
the radii of their circumscribing circles.
274
THEORETICAL GEOMETRY
BOOK V
131. Definition. — If two polygons of the same
number of sides have the angles of one taken in order
around the figure respectively equal to the angles of
the other in order, and have also the corresponding
sides in proportion, the polygons' are said to be
similar polygons.
PROBLEM 1
To describe a polygon similar to a given polygon,
and with the corresponding sides in a given ratio.
H
Let ABODE be the given polygon, and GH a st.
line taken such that AB is to GH in the given ratio.
It is required to describe on GH a polygon similar
to A BCD E and such that AB and GH are corresponding
sides.
Join AC, AD.
Make Z H = Z B, ZHGK-Z BAC and produce the
arms to meet at K. Make Z KGL = Z CAD, Z GKL =
Z ACD, and produce the arms to meet at L. Make
L LGM = Z DAE, Z GLM = Z ADE and produce the
aims to meet at M.
GHKLM is the required polygon.
Z H = Z B, Z HGK - Z BAC, .. L HKG = L BCA.
AREAS OF SIMILAR FIGURES 275
Similarly Z GLK - Z ADC, and Z M = Z E.
Hence Z HKL = Z BCD, Z KLM = Z CDE and Z HGM
= Z BAE.
.-. polygon GHKLM has its zs equal respectively to
the zs of polygon ABCDE.
r* w w i^ \/c c**
From the similar As GHK, ABC, ^ = gC = CA~ ;
and from the similar As GKL, ACD, ^- = ^;
• GH _ HK _ KL_
" AB~ = BC" ~CD~"
In the same manner it may be shown that each of
^
these ratios equals =— and .*. equals =—
Hence the corresponding sides of the two polygons
are proportional; /. polygon GHKLM is similar to
polygon ABCDE; and the two polygons have their
corresponding sides in the given ratio.
132.— Exercises
1. Draw diagrams to show that two quadrilaterals may
have the sides of one respectively proportional to the sides
of the other, but the zs of one not equal to the corre-
sponding zs of the other.
2. Draw diagrams to show that two quadrilaterals may
have the zs of one respectively equal to the zs of the
other, 'but the corresponding sides not in the same pro-
portion.
3. KLMN is a polygon. Construct a polygon similar to
KLMN, and having each side one-third of the corresponding
side of KLMN.
276
THEORETICAL GEOMETRY
BOOK V
4. ABODE is a given polygon and GH a given st. line.
Cut off AQ = GH. Take any point P within ABODE.
Join P to A, B, 0, D, E. Draw QK || AP, KF || AB, FN ||
AE, NM || ED, KL || BO. Join LM.
Show that FKLMN is similar to ABODE.
5. Twice as many polygons may be described on a given
st. line GH, each similar to a given polygon, as the given
polygon has sides.
PROBLEM 2
To divide similar polygons into similar triangles.
A E F L
Let ABODE, FGHKL be similar polygons of which
AB and FG are corresponding sides.
AREAS OF SIMILAR FIGURES 277
It is required' to divide ABODE, and FGHKL into
similar As.
Take any point P within the polygon ABODE. Join
PA, PB, PC, PD, PE.
Make Z GFQ = L BAP and Z FGQ = Z ABP, and let
the arms of these Zs meet at Q.
Join QH, QK, QL.
Z PAB = Z QFG and L PBA - L QGF; /. L FQG = L
APB, and consequently As ABP, FGQ are similar;
• QQ = FG
' PB~ AB'
But, by definition of similar polygons,
FG GH
A~B * BO'
. QG = GH
' PB ~ BO*
Also Z FGH = Z ABC and Z FGQ = Z ABP;
.'. Z QGH = Z PBC.
Then in As QGH, PBC, 5^ = |"f and Z QGH =
PD D O
Z PBC.
/. these As are similar. (IV — 10, p. 241.)
In the same manner it may be shown that the
remaining pairs of corresponding As are similar.
278 THEORETICAL GEOMETRY BOOK V
THEOREM 2
The areas of similar polygons are proportional to
the squares on corresponding sides.
A E F L
Using the diagram and construction of Problem 2
L polygon FGHKL FG2
It is required to show that *-/-? = _.
polygon ABODE AB2
V As FGQ, ABP are similar,
A FGQ = GQ2
A ABP BP2'
(V-l, p. 271.)
Similarly
BP2
A QGF _ A QGH _
A QHK A QKL _ A QLF
A PCD ~ A PDE ~ A PEA*
But, if any number of fractions be equal to each
other, the sum of their numerators divided by the
sum of their denominators equals each of the fractions.
Now the sum of the numerators of the equal
fractions is the polygon FGHKL, and the sum of the
denominators is the polygon ABODE;
polygon FGHKL _ A QFG
polygon ABODE A PAB
AREAS OF SIMILAR FIGURES 279
But A QFG - FQ2 -
Ut A PAB ~~ AB*
polygon FGHKL FG2
polygon ABODE ~ AB2"
THEOREM 3
If three straight lines are in continued proportion,
the first is to the third as any polygon on the first
is to the similar and similarly described polygon on
the second.
Hypothesis. — AB, CD, E are three st. lines such that
AB : CD = CD : E, and L, M, similar polygons having
AB, CD corresponding sides.
To prove that polygon L : polygon M = AB : E.
Proof.- o^on L = AB! (V-2, p. 278.)
Polygon M CD*
AB AB
C5D* C5D
AB CD
AB
E
280
THEORETICAL GEOMETRY
BOOK V
PROBLEM 3
To make a polygon similar to a given polygon
and such that their areas are in a given ratio.
H
Let ABODE be the given polygon and FG, GH two
given st. lines.
It is required to make a polygon similar to ABODE,
and such that its area is to that of ABODE as GH is
to FG.
Construction. — Find KL a fourth proportional to FG,
GH, AB. (IV— Prob. 2, p. 227.)
Find KM a mean proportional to FK, KL. (IV — Prob.
5, p. 246.)
Cut off AN = KM, and on AN construct a polygon
ANOPQ similar to ABODE.
AREAS OF SIMILAR FIGURES 281
.
... Polygon ABODE AB 279
polygon ANOPQ KL
~ GH
And /. Polyg°n ANOPQ _ GH
polygon ABODE FG"
"282
THEORETICAL GEOMETRY
BOOK V
PROBLEM 4
\
To make a figure equal to one given rectilineal
figure and similar to another.
M
Let D and EFGH be the given figures.
It is required to make a figure similar to EFGH
and equal to D.
Construction. — Construct the rect. KL = D, and the
rect. FN = EFGH.
Make KM the side of a square which is equal to
KL, and FO the side of a square which is equal to
FN ; so that, KM2 = D and FO2 = EFGH.
From F draw a st. line FQ and from it cut off
FP = KM and FQ = FO.
Join QE, and draw PR || QE cutting EF at R.
On RF describe RFTS similar to EFGH.
RFTS is the required figure.
EXERCISES 283
Proof.— V RFTS III EFGH,
i£ <V-1, P. 271.)
PF2
QR (IV~2> P- 222->
KM2 D
~ FO2 " EFGH
RFTS D
•"• EFGH ~~ EFGH '
and /. RFTS = D ;
also RFTS was made similar to EFGH.
133. — Exercises
1. On a plan of which the scale is 1 inch to 2 feet, a
room is represented by 30 sq. in. Find the area of the
room.
2. On a map of which the scale is 4 inches to the mile,
a farm is represented by 10 sq. in. Find the number of
acres in the farm.
3. Construct an equilateral A equal in area to a given
square.
4. Construct a square equal in area to a given A.
5. Construct a rectangle similar to a given rectangle and
equal in area to a given square.
6. Construct a square the area of which is 15 sq. in.
7. Bisect a given A by a st. line drawn _L to one side.
284
THEORETICAL GEOMETRY
BOOK V
ARCS AND ANGLES
134. — Suppose an angle AOB at the centre of a circle
to be divided into a number
of equal parts AOC, COD, DOE,
EOB.
Then, by III— 13, p. 167, the
arcs AC, CD, DE, EB are equal
to each other, and whatever mul-
tiple the angle AOB is of the
angle AOC, the arc AB is the
same multiple of the arc AC.
Thus, if an angle at the centre of a circle be
divided into degrees and contain a of them, the arc
subtending the angle will contain the arc subtending
one degree a times.
THEOREM 4
In equal circles, angles, whether at the centres
or circumferences, are proportional to the arcs on
which they stand.
Hypothesis. — In the equal circles AEB, CFD, the z_s
AOB, CQD at the centres stand respectively on the
arcs AB, CD.
ANALYSIS OF A PROBLEM — TANGENTS OF CIRCLES 285
'L AOB arc AB
To prove that - =
L CQD arc CD
Proof. — Let the zs AOB, CQD be commensurable
having Z AOH a common measure. Suppose L AOB
contains L AOH a times, and Z CQD contains Z AOH
6 times.
Then arc AB contains arc AH a times, and arc CD
contains arc AH b times.
L AOB a x L AOH a
And
L CQD b x L AOH b
arc AB _ a x arc AH _ a
arc CD b x arc AH b
L AOB arc AB
L CQD arc CD
Again, since the z_s at the circumferences are
respectively half the z_s at the centres, on the same
arcs, the /_s at the circumferences are also in the
ratio of the arcs on which they stand.
ANALYSIS OF A PROBLEM— COMMON TANGENTS OF
CIRCLES
135. A common method of discovering the solution
of a problem begins with the drawing of the given
figure or figures. The required part is then sketched
in, and a careful examination is made to determine
the connection between the given parts and the
required result. Properties of the figure are noted,
and lines are drawn that may help in finding the
solution. This method of attack is known as the
Analysis of the Problem. Its use is illustrated in
the following sections.
286
THEORETICAL GEOMETRY
BOOK V
136. Problem.— To draw the direct common tan-
gents to two given circles.
A
Let ABC, DEF be two circles, with centres P, Q.
It is required to draw a direct common tangent to
the circles ABC, DEF.
Suppose AD to be a direct common tangent touching
the circles at A, D.
Join PA, QD.
PA, QD are both J_ AD, and .*. PA || QD.
Cut off AG = DQ. Join QG.
AG is both = and |{ QD, .". AQ is a |!gm, and as L
GAD is a rt. L, AQ is a rect.
Draw a circle with centre P and radius PG.
PGQ is a rt. Z_, .*. QG is a tangent to the circle
GHK and this tangent is drawn from the given point
Q. The radius PG of the circle GHK is the difference
of the radii of the given circles.
Using the construction suggested by the above
analysis the pupil should make the direct drawing
and prove that it is correct.
Show that two direct common tangents may be
drawn.
ANALYSIS OF A PROBLEM — TANGENTS OF CIRCLES 287
137. Problem.— To draw the transverse common
tangents to two given circles.
Let ABC, DEF be two circles with centres P, Q.
It is required to draw a transverse common tangent
to the circles ABC, DEF.
Suppose AD to be a transverse common tangent
touching the circles at A, D
Join PA, QD.
PA, QD are both J_ AD, .'. PA || QD.
Produce PA to G making AG = DQ. Join QG.
Then AQ is seen to be a rect., and if a circle be
drawn with centre P and radius PG, QG is seen to be
a tangent to this circle. ,The radius PG of the circle
GHK is the sum of the radii of the given circles.
From this analysis the pupil can make the direct
construction and give the proof.
Two transverse common tangents may be drawn to
the given circles.
288 THEORETICAL GEOMETRY BOOK V
138.— Exercises
1. Draw diagrams to show that the number of common
tangents to two circles may be 4, 3, 2, 1 or 0.
2. Draw a st. line to cut two given circles so that the
chords intercepted on the line may be equal respectively to
two given st. lines.
3. P, Q are the centres of two circles. A common
tangent (either direct or transversal) meets the line of
centres at R. Show that the ratio PR : QR equals the
ratio of the radii of the circles.
4. The transverse common tangents and the line of
centres of two circles are concurrent.
5. The direct common tangents and the line of centres
of two circles are concurrent.
6. P, Q are the centres of two circles and PA, QB any
two || radii drawn in the same direction from P, Q. Show
that AB produced and the direct common tangents meet
the line of centres at the same point.
7. P, Q are the centres of two circles and PA, QB any
two || radii drawn in opposite directions from P, Q. Show
that AB and the transverse common tangents meet the line
of centres at the same point.
8. Draw the direct common tangents to two equal circles.
MISCELLANEOUS EXERCISES 289
Miscellaneous Exercises
1. Draw four circles each of radius If inches, touching
a fixed circle of radius 1 inch and also touching a st. line
1J inches distant from the centre of the circle.
2. DE, FG are |J chords of the circle DEGF. Prove that
DE.FG = DG2 - DF2.
3. If two circles touch externally at A and are touched
at B, C by a st. line, the st. line BC subtends a rt. L
at A.
4. Of all As of given base and vertical L , the isosceles
A has the greatest area.
5. ABC is an equilateral A inscribed in a circle, P is
any point on the circumference. Of the three st. lines
PA, PB, PC, shew that one equals the sum of the other
two.
6. Construct a rt.- L d A, given the radius of the
inscribed circle and an acute L of the A.
7. The diagonals AC, BD of a cyclic quadrilateral A BCD
cut at E. Show that the tangent at E to the circle circum-
scribed about A ABE is || to CD.
8. A, B, C are three points on a circle. The bisector
of L ABC meets the circle again at D. DE is drawn || to
AB and meets the circle again at E. Show that DE = BC.
9. The side of an equilateral A circumscribed about a
circle is double the side of the equilateral A inscribed in
the same circle.
10. AB is the diameter of a circle and CD a chord. EF
is the projection of AB on CD. Show that CE = DF.
11. Construct an isosceles A, given the base and the
radius of the inscribed circle.
290 THEORETICAL GEOMETRY BOOK V
12. Two circles touch externally. Find the locus of the
points from which tangents drawn to the circles are equal
to each other.
13. Two circles, centres C, D, intersect at A, B. PAQ
is a st. line cutting the circles at P, Q. PC, QD intersect
at R. Find the locus of R.
14. Two circles touch internally at A; BC, a chord of
the outer circle, touches the inner circle at D. Show that
AD bisects L BAG.
15. P is a given point on the circumference of a circle,
of which AB is a given chord. Through P draw a chord
PQ that is bisected by AB.
16. On a given base construct a A having given the
vertical Z and the ratio of the two sides.
17. AB is a given st. line and P, Q are two points such
that AP:PB = AQ:QB. Show that the bisectors of .L s
APB, AQB cut AB at the same point.
18. AB is a given st. line and P, Q are two points such
that AP : PB = AQ : QB. Show that the bisectors of the
exterior L. s at P, Q of the As APB, AQB meet AB
produced at the same point.
19. AB is a given st. line and P is a point which moves
so that the ratio AP : PB is constant. The bisectors of
the interior and exterior _ s at P of the A APB, meet AB
and AB produced at C, D respectively. Show that the
locus of P is a circle on CD as diameter.
20. AB is a st. line 2 inches in length. P is a point
such that AP is twice BP. Construct the locus of P.
21. Two circles touch externally, and A, B are the points
of contact of a common tangent. Show that AB is a mean
proportional between their diameters.
MISCELLANEOUS EXERCISES 291
22. If on equal chords segments of circles be described
containing equal L s, the circles are equal.
23. Construct a quadrilateral such that the bisectors of
the opposite L s meet on the diagonals.
24. Draw a circle to pass through a given point and
touch two given st. lines.
25. Draw a circle to touch a given circle and two given
st. lines.
26. Draw a circle to pass through two given points and
touch a given circle.
27. Construct a rt.- /_ d A given the hypotenuse and the
radius of the inscribed circle.
28. In A ABC the inscribed circle touches AB, AC at
D, E respectively. The line joining A to the centre cuts
the circle at F. Show that F is the centre of the inscribed
circle of A ADE.
29. The inscribed circle of the rt.- L d A ABC touches the
hypotenuse BC at D. Show that rect. BD.DC = A ABC.
30. If on the sides of any A equilateral As be described
outwardly, the centres of the circumscribed circles of the
three equilateral As are the vertices of an equilateral A.
31. Describe three circles to touch each other externally
and a given circle internally.
32. Show that two circles can be described with the
middle point of the hypotenuse of a rt.- L d A as centre to
touch the two circles described on the two sides as
diameters.
33. A st. line AB of fixed length moves so as to be
constantly || to a given st. line arid A to be on the
circumference of a given circle. Show that the locus of B
is an equal circle.
-292 THEORETICAL GEOMETRY BOOK V
34. Construct an isosceles A equal in area to a given A
and having the vertical / equal to one of 'the /_s of .the
given A.
35. If two chords AB, AC, drawn from a point A in the
circumference of the circle ABC, be produced to meet the
tangent at the other extremity of the diameter through A
in D, E respectively, then the A A ED is similar to A ABC.
36. If a st. line be divided into two parts, the sq. on
the st. line equals the sum of the rectangles contained by
the st. line and the two parts.
37. ABCD is a quadrilateral inscribed in a circle. AB, DC
meet at E and BC, AD meet at F. Show that the sq. on
EF equals the sum of the sqs. on the tangents drawn from
E, F to the circle.
38. The st. line AB is divided at C so that AC = 3 CB.
Circles are described on AC, CB as diameters and u
common tangent meets AB produced at D. Show that BD
equals the radius of the smaller circle.
39. DE is a diameter of a circle and A is any point on
the circumference. The tangent at A meets the tangents
at D, E at B, C respectively. BE, CD meet at F. Show
that AF is || to BD.
40. TA, TB are tangents to a circle of which C is the
centre. AD is _L BC. Show that TB : BC = BD : DA.
41. ABCD is a quadrilateral inscribed in a circle. BA, CD
produced meet at P, and AD, BC produced meet at Q.
Show that PC : PB = QA : QB.
42. Divide a given arc of a circle into two parts, so that
the chords of these parts shall be to each other in the
ratio of two given st. lines.
43. Describe a circle to pass through a given point and
touch a given st. line and a given circle.
MISCELLANEOUS EXERCISES 293
44. LMN is a rt.-lid A with L the rt. L . On the three
sides equilateral As LEM, MFN, NDL are described out-
wardly. LG is J_ MN. Prove that A FGM = A LEM and
A FGN - A NDL.
45. L is the rt. L of a rt.- L d A LMN in which LN =
2 LM. Also LX J_ MN. Prove that LX = £ MN.
46. A st. line meets two intersecting circles in P and Q,
R and S and their common chord in O. Prove that OP,
OQ, OR, OS, taken in a certain order, are proportionals.
47. LMN is a semi-circle of which O is the centre, and
OM _L LN: A chord LDE cuts OM at D. Prove that LM
is a tangent to the circle MDE.
48. The bisector of L F of A FGH meets the base GH
in E and the circumcircle in D. Prove that DG2 = DE.DF.
49. POQ, ROS are two st. lines such that PO : OQ = 3 : 4
and RO : OS = 2 : 5. Compare areas of As POR, QOS ; and
also areas of AS POS, QOR.
50. Trisect a given square by st. lines drawn || to one
of its diagonals.
51. Construct a A having its base 8 cm., the other sides
in the ratio of 3 to 2, and the vertical L - 75°.
52. In two similar As, the parts lying within the A of
the right bisectors of corresponding sides have the same
ratio as the corresponding sides of the A.
53. KMN, LMN are As on the same base and between
the same ||s. KN, LM cut at E. A line through E, || MN,
meets KM in F and LN in G. Prove that FE = EG.
54. Construct a A having given the vertical L. , the
ratio of the sides containing that L. , and the altitude
drawn to the base.
294 THEORETICAL GEOMETRY BOOK V
55. From a point P without a circle two secants PFG,
RED are drawn, and PQ drawn || FD meets GE produced
at Q. Prove that PQ is a mean proportional between
QE, QG.
56. LD bisects L L of A LMN and meets MN at D.
From D the line DE || LM meets LN at E, and DF || LN
meets LM at F. Prove that FM : EN = LM2 : LN2.
57. LMN is a A /. rt.-d at L. LD JL MN and meets a
line drawn from M _J_ LM at E. Prove that A LMD is a
mean proportional between As LDN, MDE.
58. Two circles touch externally at D and PQ is a common
tangent. PD and QD produced meet the circumferences at
L, M respectively. Show that PM and QL are diameters of
the circles.
59. The common tangent to two circles which intersect
subtends supplementary L s at the points of intersection.
60. Two circles intersect at Q and R, and ST is a
common tangent. Show that the circles described about As
STR, STQ are equal.
61. A st. line DEF is drawn from D the extremity of a
diameter of a circle cutting the circumference at E and a
fixed st. line _L to the diameter at F. Show that the rect.
DE.DF is constant for all positions of DEF.
62. A chord LM of a circle is produced to E such that
ME is one-third of LIVi ; a tangent EP is drawn to the
circle and produced to D such that PD = EP. Prove that
A ELD is isosceles.
63. Draw a st. line to touch one circle arid to cut another,
the chord cut off being equal to a given st. line.
MISCELLANEOUS EXERCISES 295
64. Two equal circles are placed so that the transverse
common tangent is equal to the radius. Show that the
tangent from the centre of one circle to the other equals
the diameter of each circle.
65. Construct a A having its medians respectively equal
to three given st. lines.
66. Construct a A given one side and the lengths of the
medians drawn from the ends of that side.
67. Construct a A given one side, the median drawn to
the middle point of that side, and a median drawn from
one end of that side.
68. Construct a A having Z A = 20°, Z C = 90°, and
c - a = 4 cm.
69. Construct a A having L C = 90°, 6 = 6 cm., and
c - a = 3*5 cm.
70. Construct a A having a = 7 cm., c - b = 3 cm., and
L C - Z B = 28°.
71. If a st. line be drawn, in any direction from one vertex
i.f a || gm, the J_ to it from the opposite vertex equals the
sum or difference of the JLs to it from the two remaining
vertices.
72. PQ is a chord of a circle JL to the diameter LM,
and E is any point in LM. If PE, QE meet the circum-
ference in S, R respectively, show that PS = Q-R; and that
RS _L LM.
73. P is any point in a diameter LM of a circle, and
QR is a chord || LM. Prove that PQ2 + PR2 = PL2 + PM2.
74. On the hypotenuse EF of the rt.- /. d A DEF a A
is described outwardly having ZGEF = ZDEF and ZGFE a
rt. L. . Prove that A GFE : A DEF = GE : ED.
296 THEORETICAL GEOMETRY BOOK V
75. Two quadrilaterals whose diagonals intersect at equal
/ s are to one another in the ratio of the rectangles contained
by the diagonals.
76. P is any point in the side LM of a A LMN. The st.
line MQ, || PN , meets LN produced at Q ; and X, Y are points
in LM, LQ respectively, such that LX2 = LP.LM and LY2 =
LN.LQ. Prove that A LXY = A LMN.
77. EFP, EFQ are circles and PFQ is a st. line. ER is a
diameter of circle EFP and ES a diameter of EFQ. Prove
A EPR : A EQS as the squares on the radii of the circles.
78. If P is the point of intersection of an external common
tangent PQR to two circles with the line of centres, prove
that PQ : PR as the radii of the circles. Also, if PCDEF is
a secant, prove that PC:PE = PD:PF
79. A point E is taken within a quadrilateral FGHK such
that Z.EFK = ZGFH and AEKF = Z.GHF. GE is joined.
Prove A PEG ||| A FHK.
80. Through a given point within a circle, draw a chord
that is divided at the point in a given ratio
81. From P, a point on the circumference of a circle,
tangents PE, PF are drawn to an inner concentric circle.
GEFH is a chord, and PE meets the circumference at Q.
Prove As PGF, PEH, GEQ similar; also show that GQ2:
GP2 = GE : GF.
82. L is the vertex of an isosceles A LMN inscribed in
a circle, LRS is a st. line which cuts the base in R and
meets the circle in S. Prove that SL. RL = LM2.
83. PQR is a rt.-/d A with P the rt. L. PD J_ QR ;
DM J_ PQ and DN J_ PR. Prove that L QMR = L QNR.
84. DEF is an isosceles A with L D = 120.° Show that
if EF be trisected at G and H, the A DGH is equilateral.
MISCELLANEOUS EXERCISES 297
85. AS and AT, BP and BQ are tangents from two
points A and B to a circle. C, D, E, F are the middle
points of AS, AT, BP, BQ respectively. Prove that CD,
EF, produced if necessary, meet on the right bisector of AB.
(Let O be the centre of the circle; L and M the points where
OA, OB cut the chords of contact. Prove A, L, M, B con-
cyclic, etc.)
86. If from the middle point of an arc two st. lines be
drawn cutting the chord of the arc and the circumference,
the four points of intersection are concyclic.
87. If a st. line be divided at two given points, find a
third point in the line, such that its distances from the
ends of the line may be proportional to its distances from
the two given points.
88. Prove geometrically that the arithmetic mean between
two given st. lines is greater than the geometric mean
between the two st. lines.
89. A square is inscribed in a rt. -angled triangle, one
side of the square coinciding with the hypotenuse : prove
that the area of the square is equal to the rectangle con-
tained by the extreme segments of the hypotenuse.
90. Any regular polygon inscribed in a circle is the
geometric mean between the inscribed and circumscribed
regular polygons of half the number of sides.
91. The diagonal and the diagonals of the complements
of the parallelograms about the diagonal of a parallelogram
are concurrent.
92. Develop the formula for the area of a A,
\/'s(s -a) (s — b) (s — c) where 2s = a + b + c and a, b, c are
the sides.
298
THEORETICAL GEOMETRY
BOOK V
Solution of 92. In A ABC, draw AX _L BC, and let AX
= A, BX = x. Then CX = a - x.
Area of A ABC = J a h.
= c* —
(«a _ £2 + C2)2
62 + c2)2
+ c2) (2a
- a +
- 2
c2)
4a2
4a2 A2 = 4a2 c2 - (a2
= (2ac + a2 -
= {(a + o)2 -
= (a + b + c) (a - b + c) (a + b - c) (b - a + c)
= 2s (2s - 26) (2s - 2c) (2s - 2a).
| a2 A2 = s (s - a) (s - b) (s - c),
And
B
- a) (s - 6) (s - c).
93. Show from the diagram how
the distance between two points,
A, B at opposite sides of a pond
may be found by measurements
on land.
94. Show from the diagram how the breadth of a river
may be found by measurements
made on one side of it.
B
95. Given a st. line AB,
construct a continuation of it
CD, AB and CD being sepa-
rated by an obstacle.
96. AB, CD are two lines
which would meet off the
paper. Draw a st. line which would pass through the point
of intersection of AB, CD, and bisect the L between them.
INDEX TO DEFINITIONS
PAGE
Acute angle : — An L which is < a rt. L 10
Acute-angled triangle : — A A which has three acute L s 27
Adjacent angles : — Two Z s which have the same vertex, an arm
common, and the remaining arms on opposite sides of the com-
mon arm 11
Altitude of a triangle : — The length of the J_ from any vertex of
the A t° the opposite side 27
Angle : — The amount of rotation made by a st. line when it revolves
about a fixed point in itself from one position to another 8
Antecedent : — The first term in a ratio 213
Arc of a circle : — A part of the circumference 17
Axiom : — A statement that is self-evident, or assumed ,. 3
Axis of symmetry : — The line about which a symmetrical figure
can be folded so that the parts on one side will exactly fit the
corresponding parts on the other side . 21
Centroid : — The point where the medians of a A intersect one
another 69
Chord of a circle : — The st. line joining two points on the cir-
cumference 17
Chord of contact : — The st. line joining the points of contact of
two tangents 174
Circle : — The locus of the points that are at a fixed distance from a
fixed point 17,141
(The name circle is also used for the area inclosed by the circumference).
Circumcentre : — The centre of the circumscribed circle of a A ••'•• 144
Circumference : — Same as circle 17
Circumscribed circle :— A circle which passes through all the ver-
tices of a rectilineal figure 143
Coincide :— Magnitudes which fill exactly the same space are said
to coincide with each other 4
Commensurable magnitudes : — Magnitudes which have a common
measure 214
Complementary angles : — Two La of which the sum is one rt. L . 13
Concyclic points : — Points through which a circle may be described 143
Congruent figures :— Figures equal in all respects, so that one may
be made to fit the other exactly 15
Consequent : — The second term of a ratio 213
Converse propositions :— Two propositions of which the hypothesis
of each is the conclusion of the other 41
299
300 INDEX TO DEFINITIONS
PAGE
Cyclic quadrilateral : — A quadrilateral of which all the vertices
are on the circumference of the same circle « 144
Diagonal :— A st. line joining opposite vertices of a rectilineal
figure 17
Diameter of a circle : — A chord passing through the centre 17
Equilateral triangle : — A A having three equal sides 19
Escribed circle of a triangle : —A circle which is touched by one
side of a A an<l by the other two sides produced 187
External segments of a line : — The distances from the two ends of
a line to a point taken in the line produced 217
Figure : — Any combination of points, lines, surfaces and solids ... 2
Geometric mean : — The second of three terms that are in continued
proportion 214
Hypotenuse : — The side opposite the rt. Z in a rt.- Z d A 27
Hypothesis : — The part of a proposition in which the conditions
that are supposed to exist are stated 5
Incommensurable magnitudes : — Magnitudes that have no com-
mon measure however small 214
Indirect method of demonstration : — Method of proof that begins
by assuming that the proposition to be proved is not true 41
Inscribed Circle : — A circle that is within a closed, rectilineal
figure, and which is touched by each side of the figure 187
Internal segments of a line : — The distances from the ends of a
line to a point in the line 217
Isosceles triangle : — A A having two equal sides 19
Line : — That which has length, but neither breadth nor thickness.. 1
Locus :— A figure consisting of a line (or lines) which contains all
the points that satisfy a given condition, and no others 77
(See also page 78).
Major arc of a circle :— An arc that is greater than half the cir-
cumference 153
Major segment of a circle : — A segment of which the arc is a
major arc 153
Means of a proportion : — The second and third of four magnitudes
that are in proportion 214
Mean proportional : — The second of three magnitudes that are in
continued proportion 214
Median ;— A st. line drawn from any vertex of a A to the middle
point of the opposite side . . 30
Minor arc of a circle :— An arc that is less than half the circum-
ference ... 153
INDEX TO DEFINITIONS 301
PAGE
Minor segment of a' circle :— A segment of which the arc is a
minor arc 153
Obtuse angle : — An L which is > a rt. L 10
Obtuse-angled triangle :— A A one L of which is an obtuse L . . 27
Parallel straight lines : — St. lines in the same plane which do not
meet when produced for any finite distance in either direction. 35
Parallelogram : — A quadrilateral that has both pairs of opposite
sides || to each other 35
Perpendicular : — Each arm of a rt. L is said to be J_ to the other
arm 10
Plane surface : — A surface such that if any two points on it be
joined by a st. line the joining line lies wholly on the surface. . 2
Point :— That which has position but no size 1
Point of contact : — The common point of a tangent and circle. . . . 169
Polygon : — A figure bounded by more than four st. lines. Also
sometimes used for a figure bounded by any number of st. lines. 68
Problem : — The statement of a construction to be made 4
Projection of the point on the line : — The foot of the _L from the
point to the line 128
Projection of the line on the line : — The intercept on the second
line between the projections of the two ends of the first line on
the second . 129
Proportion : — The equality of ratios 213
Proposition : — That which is stated or affirmed for discussion 4
Quadrilateral : — A closed figure formed by four st. lines 17
Radius: — A st. line drawn from the centre of a circle to the
circumference 17
Ratio :— The measure of one magnitude when another magnitude
of the same kind is taken as the unit 213
Rectangle :— A ||gm of which the Z s are rt. Z s 68
Rectilineal figure : — A figure formed by st. lines 15
Reflex angle : — An Z which is > two rt. Zs, but < four rt. Zs.. 154
Regular polygon :— A polygon in which all the sides are equal to
each other, and all the L s are equal to each other 68
Rhombus :— A quadrilateral having its four sides equal to each other 17
Right angle : — An Z which is half of a sfc. Z 10
Right-angled triangle : — A A one L of which is a rt. L 27
Right bisector : — A st. line which bisects a st. line of given length
at rt. La . . 28
Scalene triangle : — A A having no two of its sides equal to each
other . . 19
302 INDEX TO DEFINITIONS
Secant of a circle : — A st. line drawn from a point without to cut
a circle < 169
Sector of a circle: — A figure bounded by two radii of a circle and
either of the arcs intercepted by these radii . . . 155
Segment of a circle : — A figure bounded by an arc of a circle and
the chord which joins the ends of the arc 153
Similar polygons : — Two polygons of the same number of sides
which have the Z_s of one taken in order around the figure
respectively equal to the Z_s of the other in order, and have
also the corresponding sides in proportion 274
Similar segments of circles :— Segments which contain equal Z_s. 263
Similar triangles: — Two AS which have the three La of one
respectively equal to the three Z_s of the other 158
Solid :— That which has length, breadth and thickness 2
Square : — A rectangle of which all the sides are equal to each other. 68
Straight angle : — Half of a complete revolution made by a st. line
revolving about a point in itself 9
Straight lines : — Lines which cannot have any two points of one
coincide with two points of the other without the lines coin-
ciding altogether 1
Subtend : — A line drawn from a point in one arm of an L to a
point in the other arm subtends the L 17
Supplementary angles : — Two Z_s of which the sum is two rt. Z_s. 13
Surface : — That which has length and breadth but no thickness ... 2
Symmetrical figure: — A figure which can be folded along a st. line
so that the parts on one side exactly fit the corresponding parts
on the other side , 20
Tangent to a circle : — A st. line which, however far it may be
produced, has one point on the circumference of a circle and all
other points without the circle 169
Theorem : — The statement of a truth to be proved 4
Touch • — A tangent is said to touch a circle 169
Two circles which meet each other at one point and only one
point are said to touch each other 196
Transversal — A st. line which cuts two, or more, other st. lines. . 35
Triangle : — A figure formed by three st. lines which intersect one
another 15
Vertex of an angle :— The point from which the two arms of the
L are drawn 8
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YB 17296
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